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ANSWER: ANSWER: Correct Because this calculation involved division, the final answer must be rounded to the same number of significant figures as the measured value with the fewest significant figures. Part C Compute . Round the answer appropriately. Express your answer numerically using the proper number of significant figures. Hint 1. How to approach the problem The rules for addition and subtraction differ from those for multiplication and division. For addition and subtraction, the answer should be rounded to the same number of decimal places as the number having the fewest decimal places. Hint 2. Determine the number of digits How many digits to the right of the decimal point should the final answer have after rounding? Express the number of significant figures as an integer. ANSWER: ANSWER: 3 99.3 = 1240.64 12.5 9.3456 + 2140.56 2 2149.91 9.3456 + 2140.56 =

ANSWER: Correct Because this calculation involved subtraction and both values have no digits following the decimal point, the final answer must be rounded to the same number of digits as the measured value with the fewest number of significant digits to the left of the decimal point. Rounding It will be helpful to remember two guidelines about significant figures. 1. After multiplication or division, the answer cannot have more significant figures than any of the original numbers. 2. After addition or subtraction, the answer cannot have more digits to the right of the decimal point than any of the original numbers. After determining the appropriate number of significant figures, round off your final answer. 1. If the first digit you drop is greater than or equal to 5, add 1 to the last digit you keep. You are rounding up. 2. If the first digit you drop is less than 5, do nothing to the digits you keep. You are rounding down. Part A A bakery measured the mass of a whole cake as 0.840 . A customer bought one slice of the cake and measured the mass of the slice as 0.1171 . What is the mass of the remaining cake, without the one slice? Express your answer in kilograms using the appropriate number of significant figures. ones tens hundreds thousands ten thousands 2.52x10^4 4.659 × − 2.14 × = 10 4 10 4 kg kg

Correct ± Using Conversion Factors Chemistry is an experimental science in which a consistent system of measurements is used. Scientists throughout the world use the International System of Units (Système International or SI), which is based on the metric system. However, the metric system in not used on an everyday basis in the United States. It is often necessary to convert between the English system, which is used in the United States, and the SI system. One problem with the SI system is the size of the units sometimes turns out to be inconveniently large or small. Therefore it is often necessary to make conversions within the SI system. For this reason, SI units are modified through the use of prefixes when they refer to either larger or smaller quantities. Part A In Europe, gasoline efficiency is measured in . If your car's gas mileage is 24.0 , how many liters of gasoline would you need to buy to complete a 142- trip in Europe? Use the following conversions: and . Express your answer numerically in liters. Hint 1. How to approach the problem Several conversions are necessary for this problem. • First convert miles to kilometers using the conversion factor . That way the efficiency of your car can be measured in . • Next convert gallons to liters using the conversion factor . • Finally, determine how many liters of gasoline are needed to complete a 142- trip in your car. Hint 2. Convert miles to kilometers With an efficiency of 24.0 , your car can travel 24.0 on 1 of gas. How far is that in kilometers given the equality ? Express your answer numerically in kilometers. ANSWER: 1.77 g/cm 3 km/L mi/gal km 1 km = 0.6214 mi 1 gal = 3.78 L 1 km = 0.6214 mi km/gal 1 gal = 3.78 L km mi/gal mi gal 1   km = 0.6214   mi 38.6 km

to calculate the average speed. Hint 2. Convert meters to kilometers The prefix kilo means times 1000. There are 1000 meters in one kilometer. Use this information to complete the calculation. The distance the sprinter ran was 200.0 meters. How many kilometers is this? Express your answer numerically in kilometers. ANSWER: Hint 3. Convert seconds to hours To convert seconds to hours, use the following conversions: 1. and 2. . The sprinter completed the race in 21.8 . Convert that value to hours. Express your answer numerically in hours. ANSWER: ANSWER: Correct Approximately 33-35 kilometers per hour is equal to 20-22 miles per hour. This is the top speed that humans can possibly move on foot. Even the best sprinters in the world would not beat this speed by much. Answers that are much higher than this would not be possible by humans. Answers that are much lower than this would be too slow for a top athlete. Part D A specific brand of gourmet chocolate candy contains 7.00 of dietary fat in each 22.7- piece. How many kilograms of dietary fat are in a box containing 3.00 of candy? 200.0 = 0.2000 m km 60   s = 1   min 60   min = 1   h s 21.8 = 6.06 × 10 − 3 s h 33.0 km/h g g lb

1.2 Introduction - reactions and mole calculations Due: 11:59pm on Sunday, September 17, 2023 You will receive no credit for items you complete after the assignment is due. Grading Policy Chemical Properties and Physical Properties Substances have both chemical properties and physical properties . • Chemical properties describe the characteristic ability of a substance to react to form new substances through a new arrangement of atoms • Physical properties are characteristics that can be measured without changing the composition of the sample under study. The appearance, but not the composition, will change. Part A Which of the following properties of copper are chemical properties? Check all that apply. Hint 1. How to identify chemical properties Changes that result in a new arrangement of atoms and, thus, new compounds are associated with chemical properties. To identify whether a new substance was formed, look for signs of a reaction such as two reagents coming together to form one or more different products. This may involve visual differences as well; however, only chemical properties involve different substances too. ANSWER: Correct In air it combines with carbon dioxide to form copper carbonate. It reacts with chlorine to form copper chloride. It is a shiny, orange-colored metal. It has a mass of 65.55 . g

Drag the appropriate definitions to their respective words. ANSWER: ANSWER: Reset Help It is used as a disinfectant and bleach. It is a fuel that is burned to produce the energy. It is made up of tiny droplets of water suspended in air. It is the temperature below which the water vapor from the atmosphere forms tiny droplets of liquid water.

ANSWER: Correct Reset Help 1. The process by which glucose, a sugar, is converted to ethanol and carbon dioxide gas is called fermentation . 2. The burning of wood in a fireplace converts the wood to ashes, carbon dioxide, and water vapor. 3. evaporation of liquid water to water vapor causes the liquid in a glass to disappear on standing. 4. On a warm day, water vapor from the atmosphere forms liquid water on the outside of a glass by condensation . 5. When a complex material is broken down into simpler substances it is called decomposition . formation of a carbon deposit inside the flask formation of carbon dioxide gas from glucose condensation of ethanol formation of ethanol from glucose by yeast burning of natural gas evaporation of ethanol

Calculations Using the Mole Learning Goal: To learn how to convert grams to moles and to use the mole to find the number of atoms in a sample. The mole (abbreviated ) is a counting unit used to simplify calculations that would otherwise involve very large numbers. The mole is equivalent to the number of carbon atoms in exactly 12 of isotopically pure , or . This number is known as Avogadro's number in honor of Amedeo Avogadro. Avogadro's number can be used as a conversion factor between moles and atoms as shown here: The molar mass of a substance is the mass of one mole of a substance and is written in units of grams per mole. The molar mass of an atom is equivalent to its atomic mass whereas the molar mass of a substance is equivalent to its formula weight. Part A How many moles of atoms are in 6.00 of ? Express your answer numerically in moles. Hint 1. Review isotope notation A chemical isotope is written using the following notation: where is the mass number of the element with chemical symbol , and is its atomic number. Hint 2. Determine the molar mass of carbon-13 What is the molar mass of ? Express your answer numerically in grams per mole. ANSWER: Hint 3. Identify the operation that allows units to cancel properly mol g C 12 6.02 × 10 23 6.02 ×   atoms 10 23 1   mole   of   atoms g C 13 X A Z A X Z C 13 13.0 g/mol

Given of a substance that has a molar mass of , which operation will give an answer with units of moles? ANSWER: ANSWER: Correct While the value of a mole is the number of atoms of atoms in exactly 12.0 of pure , the same is not true for other elements or other isotopes of carbon. To six significant figures, the molar mass of is 13.0034 . Part B Based on your answer in Part A, calculate the number of atoms in this amount of ? Express your answer numerically in atoms of carbon. Hint 1. How to approach the problem Use Avogadro's number to convert moles of to atoms of carbon. Recall that Avogadro's number is . In other words, each mole of atoms contains atoms. ANSWER: Correct x g y g/mol x ⋅ y x y y x 0.462 mol C   13 C 12 g C 12 C 13 g/mol C 13 C 13 6.02 × 10 23 6.02 × 10 23 2.78 × 10 23 atoms C 13

Part C Based on your answer in Part B, how many electrons are in this amount of ? Express your answer numerically in electrons. Hint 1. How to approach the problem In a neutral atom, the number of protons is equal to the number of electrons. The number of protons in an atom can be determined by looking up the atomic number of the element in the periodic table . Once you know the number of electrons per atom, multiply by the total number of atoms, 2.78 × 10 23 . Hint 2. Determine the number of electrons in one atom of carbon-13 How many electrons are in one atom of ? Express your answer numerically as an integer. ANSWER: ANSWER: Correct Part D Based on your answer in Part B, how many neutrons are in this amount of ? Express your answer numerically in neutrons. Hint 1. How to approach the problem The mass number of an atom is the sum of its protons and neutrons. The number of protons in an atom can be determined by looking up the atomic number of the element in the periodic table .Once you know the number of neutrons per atom, multiply by the total number of atoms, 2.78 × 10 23 . Hint 2. Determine the number of neutrons in one atom of carbon-13 C 13 C 13 6 electrons 1.67 × 10 24 electrons C 13

How many neutrons are in one atom of ? Express your answer numerically as an integer. ANSWER: ANSWER: Correct Percent Composition and Formulas One of the inherently satisfying features of chemistry is that chemical materials form and interact in a rational and predictable manner. For example, one can predict a great deal about a particular kind of molecule by experimentally determining the percentage composition of the elements in that compound. This gives us the relative proportions of the elements in the molecule. For a molecule made up of elements , , and , the proportions might be , meaning that there are two atoms of for each atom of and each atom of . This may not be the actual description of the molecule (which might actually be ), but it is the "reduced" version of that formula, called the empirical formula . The actual formula is some multiple of the empirical formula. To know the actual formula we need to know both the empirical formula and the molecular mass of the compound. This provides us with the multiplier value in whole units that must be applied to the empirical formula to get the actual formula. A compound is 40.0 , 6.70 , and 53.3 by mass. Assume that we have a 100.- sample of this compound. Part A What are the subscripts in the empirical formula of this compound? Enter the subscripts for , , and , respectively, separated by commas (e.g., 5,6,7 ). Hint 1. Find how many moles of each element are involved To obtain the mass of each component, multiply the percentage of each component by the mass of the sample. For a 100- sample, there are 40.0 of , 6.70 of , and 53.3 of . Then, to convert grams to moles, divide the mass by the molar mass. C 13 7 neutrons 1.94 × 10 24 neutrons A B C A:B:C 2 C A B A 2 B 2 C 4 % C % H % O g C H O g g C g H g O

How many moles of each element are present in 100 of this compound? Enter the number of moles of , , and , respectively, separated by commas (e.g. 5, 6, 7 ). ANSWER: Hint 2. How to convert the mole ratio to integers To convert the ratio of moles of to the integers needed to determine the empirical formula, divide each number in the ratio by the smallest number in the ratio. ANSWER: Correct The empirical formula of this compound is . Part B The molecular formula mass of this compound is 240 . What are the subscripts in the actual molecular formula? Enter the subscripts for , , and , respectively, separated by commas (e.g., 5,6,7 ). Hint 1. How to approach the problem Based on the molecular mass of the compound, determine how many units of the empirical formula mass there are per actual mass of the molecule. The number of units is a multiplication factor ( ) that can be multiplied by each subscript in the empirical formula to give the molecular formula. Hint 2. Calculate the empirical formula mass What is the mass of the empirical formula, ? Express your answer in atomic mass units using four significant figures. ANSWER: g C H O number of moles of , , = 3.33,6.65,3.33 C H O mol C:H:O 1,2,1 C O H 2 amu C H O n C O H 2 30.03 amu

Hint 3. Equation for the multiplication factor Let be the number by which the empirical formula mass is a factor of the molecular formula mass. Find using the following equation: Express your answer as an integer. ANSWER: ANSWER: Correct Balancing Chemical Equations Learning Goal: To learn to balance chemical equations by inspection. According to the law of conservation of mass, matter cannot be created or destroyed. Therefore, a chemical equation must show the same number of each kind of atom in the reactants as it does in the products. As shown in the figure, the balanced equation has 6 atoms of hydrogen and 2 atoms of nitrogen on each side of the arrow. One is often presented with unbalanced chemical equations for which one must supply the coefficients. n n n n = molecular formula mass empirical formula mass = 8 n 8,16,8 3 + → 2N H 2 N 2 H 3

The following equation is not balanced: Part A Notice that " " appears in two different places in this chemical equation. is a polyatomic ion called "sulfate." What number should be placed in front of to give the same total number of sulfate ions on each side of the equation? Express your answer numerically as an integer. Hint 1. Confused by parentheses? Parentheses surrounding a set of elements means that the subscript applies to all the atoms within the parentheses. For example: the compound contains two aluminum atoms and three sulfate ions (12 oxygen atoms, 3 sulfur atoms). ANSWER: Correct Part B Now that we have put a coefficient of 3 in front of , what coefficient should go in front of to balance calcium ( )? Express your answer numerically as an integer. Hint 1. Count the number of calcium atoms on the left How many calcium atoms currently appear on the left side of this equation? Express your answer numerically as an integer. ANSWER: CaS + AlC → CaC + A (S O 4 l 3 l 2 l 2 O 4 ) 3 SO 4 SO 4 2 − CaSO 4 ?CaS + AlC → CaC + A (S O 4 l 3 l 2 l 2 O 4 ) 3 A (S l 2 O 4 ) 3 3 ? = CaSO 4 CaCl 2 Ca 3CaS + AlC → ?CaC + A (S O 4 l 3 l 2 l 2 O 4 ) 3 3CaS + AlC → CaC + A (S O 4 l 3 l 2 l 2 O 4 ) 3

ANSWER: Correct By placing a coefficient of 3 in front of calcium chloride, we have indicated three calcium atoms on the right to match the three calcium ions on the left: Part C Now that we have put coefficients of 3 in front of and , what coefficient should go in front of to balance both the atoms and the atoms? Express your answer numerically as an integer. ANSWER: Correct There are now 6 chlorine atoms, and 2 aluminum atoms on each side of the equation. The final balanced equation looks like this: Score Summary: Your score on this assignment is 99.4%. You received 4.97 out of a possible total of 5 points. 3 3 ? = 3CaS + AlC → 3CaC + A (S O 4 l 3 l 2 l 2 O 4 ) 3 CaSO 4 CaCl 2 AlCl 3 Cl Al 3CaS +?AlC → 3CaC + A (S O 4 l 3 l 2 l 2 O 4 ) 3 2 ? = 3CaS + 2AlC → 3CaC + A (S O 4 l 3 l 2 l 2 O 4 ) 3

2.1 Nomenclature Due: 11:59pm on Sunday, September 24, 2023 You will receive no credit for items you complete after the assignment is due. Grading Policy Ionic Compound Nomenclature and Formulas Learning Goal: To learn to write the systematic name of ionic compounds given the chemical formula and to write the chemical formula given the systematic name. Ionic bonds form when one atom completely transfers one or more electrons to another atom, resulting in the formation of ions. Positively charged ions (cations) are strongly attracted to negatively charged ions (anions) by electrical forces. All chemical compounds can be named systematically by following a series of rules. Binary ionic compounds are named by identifying first the positive ion and then the negative ion. Naming compounds with polyatomic ions involves memorizing the names and formulas of the most common ones. Part A Give the systematic name for the compound . Spell out the full name of the compound. Hint 1. How to approach the problem When naming ionic compounds, the cation is identified first and the anion is identified second. Here, the anion is a polyatomic ion. It is generally necessary to memorize the names of polyatomic ions such as (hydroxide), (ammonium), or (sulfite). However, there are a few guidelines: • The names of most polyatomic ions end in ite or ate . • Oxoanions are ions in which a given element combines with different numbers of oxygen atoms. For example, sulfur can combine with three or four oxygen atoms, forming or . When there are only two oxoanions, the ion with fewer oxygen atoms has the ending ite , and the ion with more oxygen atoms has the ending ate . The first part of the name is taken from the name of the element that combines with the oxygen atoms. Thus, is the sulfite ion and is the sulfate ion. Nitrogen forms two oxoanions, and . Hint 2. Name the anion What is the name of the polyatomic anion ? Spell out the full name of the compound. ANSWER: Al(NO 3 ) 3 OH − NH 4 + SO 3 2 − SO 3 2 − SO 4 2 − SO 3 2 − SO 4 2 − NO 2 − NO 3 − NO 3 −

ANSWER: Correct Aluminum is a metal that has only one oxidation state. Therefore it is not necessary to write III in parentheses in the systematic name. Only those metals with more than one oxidation state must have its state written in the name. Part B Give the systematic name for the compound . Spell out the full name of the compound. Hint 1. How to approach the problem When naming ionic compounds, the cation is identified first and the anion is identified second. If the metal has more than one oxidation state, identify the oxidation state using a Roman numeral in parentheses immediately after the metal name. The, when naming the anion for this part take into account that it is a polyatomic ion. It is generally necessary to memorize the names of polyatomic ions such as (hydroxide), (ammonium), or (sulfite). However, there are a few guidelines: • The names of most polyatomic ions end in ite or ate . • Oxoanions are ions in which a given element combines with different numbers of oxygen atoms. For example, nitrogen can combine with two or three oxygen atoms, forming or . When there are only two oxoanions, the ion with fewer oxygen atoms has the ending ite , and the ion with more oxygen atoms has the ending ate . The first part of the name is taken from the name of the element that combines with the oxygen atoms. Thus, is the nitrite ion and is the nitrate ion. Sulfur forms two oxoanions, and . Hint 2. Name the anion What is the systematic name of the anion ? Spell out the full name of the compound. ANSWER: nitrate aluminum nitrate N (S i 2 O 4 ) 3 OH − NH 4 + SO 3 2 − NO 2 − NO 3 − NO 2 − NO 3 − SO 3 2 − SO 4 2 − SO 4 2 −

Hint 3. Determine the oxidation number What is the charge of the nickel cation in this ionic compound? ANSWER: ANSWER: Correct Although it seems like a small difference, nickel(II) and nickel(III) behave much differently chemically. They even form different-colored compounds because of the number of electrons they have to bond. Therefore it is very important to specify which oxidation state is being used. Part C Enter the formula for the compound barium oxide. Express your answer as a chemical formula. sulfate 3 − 2 − 0 +1 +2 +3 nickel(III) sulfate

Hint 1. How to approach the problem Use the periodic table to determine the charge for each ion. Because this is a neutral ionic compound, the positive charges must equal the negative charges. Reduce the subscript on each ion to its lowest whole-number ratio. Hint 2. Determine the charge of the anion Oxygen is in group 16 on the periodic table . This group is also known as group 6A using the U.S. system. Main group nonmetals usually form anions whose charge is equal to the U.S. system group number minus eight. What is the charge of the oxide ion? ANSWER: Hint 3. Determine the charge of the cation What is the charge of the barium ion? ANSWER: 4 − 3 − 2 − 1 − 0 +1 +2 +3

ANSWER: Correct All of the elements in group 2 form ions with a +2 charge. That is because all of these elements need to lose two electrons to gain stability. After losing the electrons, there is an excess of two protons, which results in a +2 charge. Part D Enter the formula for the compound cobalt(II) phosphate. Express your answer as a chemical formula. Hint 1. How to approach the problem The oxidation number of the cation is given. The anion is polyatomic, so you will have to look up its formula and charge. Because this is a neutral ionic compound, the positive charges must equal the negative charges. Reduce the subscript on each ion to its lowest whole-number ratio. Hint 2. Determine the formula for the phosphate ion What is the formula for the phosphate anion? ANSWER: 2 − 1 − 0 +1 +2 +3 +4 BaO

Hint 3. Balance the charges Because this is a neutral ionic compound, the total number of positive charges [from the cobalt(II) ion] must equal the total number of negative charges (from the phosphate ion). The cobalt(II) ion has a +2 charge and the phosphate ion has a 3 charge. Therefore, if there were only one cobalt(II) ion and one phosphate ion, the overall charge would be 1, because . To have an overall charge of 0, this compound will have more than one of each ion. Cobalt(II) has a charge of +2 and phosphate has a charge of 3. Therefore, the cobalt(II) phosphate formula must have _____ cobalt(II) ions and _____ phosphate ions for the charges to balance. ANSWER: ANSWER: Correct Cobalt, like iron, has more than one oxidation state, so it is necessary to put the oxidation state of the metal in parentheses. Elements in group 1 and 2 only have one oxidation state and so there is no need for Roman numerals in their systematic name. P 3 − PO 4 3 − PO 3 3 − − − +2 + ( − 3) = − 1 − one / one one / two one / three two / one two / three three / two Co 3 (P ) O 4 2

Covalent Compound Formulas The chemical formula of covalent compounds can be easily determined from the systematic name of the compound. Greek prefixes are used to indicate the number of atoms of each element present in the compound. Number Prefix 1 mono 2 di 3 tri 4 tetra 5 penta 6 hexa 7 hepta 8 octa 9 nona 10 deca Part A What is the formula for the compound nitrogen monoxide? Express your answer as a chemical formula. Hint 1. Specify the symbol for nitrogen What is the elemental symbol for nitrogen? Express your answer as a chemical symbol. ANSWER: Hint 2. Specify the symbol for oxygen What is the elemental symbol for oxygen? N

Express your answer as a chemical symbol. ANSWER: Hint 3. Determine the number of oxygen atoms How many oxygen atoms should be shown in the formula for nitrogen monoxide? ANSWER: ANSWER: Correct Part B What is the formula for the compound tetraphosphorus decoxide? Express your answer as a chemical formula. Hint 1. Specify the symbol for phosphorus What is the elemental symbol for phosphorus? Express your answer as a chemical symbol. ANSWER: Hint 2. Determine the number of phosphorus atoms How many phosphorus atoms should be shown in the formula for tetraphosphorus decoxide? O 1 NO P

ANSWER: Hint 3. Specify the symbol for oxygen What is the elemental symbol for oxygen? Express your answer as a chemical symbol. ANSWER: Hint 4. Determine the number of oxygen atoms How many oxygen atoms should be shown in the formula for tetraphosphorus decoxide? ANSWER: ANSWER: Correct Part C What is the formula for the compound methane? Express your answer as a chemical formula. Hint 1. How to approach the problem Methane could also be called carbon tetrahydride or natural gas. 4 O 10 P 4 O 10

ANSWER: Correct Naming Binary Molecular Compounds Molecular compounds are usually composed solely of nonmetals. A binary molecular compound is one in which the compound contains only two elements (regardless of how many atoms are present of each). When naming binary molecular compounds, prefixes are used to specify the number of atoms of each element. Take a moment to review some of the prefixes shown here. Prefix Number mono one di two tri three tetra four penta five hexa six hepta seven octa eight nona nine deca ten For example, is named sulfur hexafluoride. Note that the prefix mono is not used in naming the first element. Also note that the second element in the name should end with the suffix ide . Part A Using the rules for naming molecular compounds described in the introduction, what is the name for the compound ? Spell out the full name of the compound. CH 4 SF 6 NO 2

Hint 1. Identify the number of atoms How many nitrogen atoms are in a molecule of ? How many oxygen atoms are there? Enter the number of nitrogen atoms followed by the number of oxygen atoms separated by a comma. ANSWER: Hint 2. Identify the prefix for oxide In naming this compound, what prefix would be used to specify the number of oxygen atoms per molecule? Enter the prefix. ANSWER: ANSWER: Correct The prefix mono- is not included on the first element, however if there were two or more nitrogen atoms, then the appropriate prefix would be necessary. Part B Using the rules for naming molecular compounds described in the introduction, what is the name for the compound ? Spell out the full name of the compound. Hint 1. Identify the number of atoms How many nitrogen atoms are in a molecule of ? How many hydrogen atoms are there? Enter the number of nitrogen atoms first followed by a comma and then the number of hydrogen atoms. NO 2 1,2 Number of nitrogen atoms, oxygen atoms di nitrogen dioxide N 2 H 4 N 2 H 4

ANSWER: Hint 2. Identify the prefix for nitrogen In naming this compound, what prefix would be used to specify the number of nitrogen atoms per molecule? Enter the prefix. ANSWER: Hint 3. Identify the prefix for In naming this compound, what prefix would be used to specify the number of hydrogen atoms per molecule? Enter the prefix. ANSWER: ANSWER: Correct Since there are 2 atoms of nitrogen in the molecule, the prefix di is necessary. Acid Names Acids are compounds that have at least one ion that can dissociate from the compound in solution. Precise naming of acids is important because different acids have very 2,4 di H tetra dinitrogen tetrahydride H +

different properties. For example, acetic acid in dilute form is used in cooking (as vinegar), whereas hydrofluoric acid is so toxic that even splashing a small amount on the skin can be fatal without proper treatment. Naming binary acids When naming acids, start by naming the anion. A monatomic anion, such as , is simple named by replacing the ending of the element name with ide . For example, is named chloride and is named oxide. Once the anion is named, name the acid based on this anion. Start by adding the prefix hydro , and then change the ide suffix to ic . For example, would require the anion chloride to be changed to chloric. Adding the prefix hydro creates hydrochloric acid. Part A What is the name of the acid whose formula is ? Spell out the full name of the acid. Hint 1. How to approach the problem To solve this problem, look at the chemical formula and determine which anion the hydrogen ion is bound to. This will allow you to distinguish between the naming convention for acids with the hydrogen ion bound to oxygen and that for other acids such as binary acids. Hint 2. Identify the anion What is the name of the anion to which the hydrogen ion is bound? Enter the name of the anion (e.g., sulfide). ANSWER: Hint 3. Naming conventions When a hydrogen ion in acid is bound to an anion whose name ends in ide , the acid is named with the prefix hydro , followed by the name of the anion, but with the suffix ic instead of ide . ANSWER: Cl − Cl − O 2 − HCl HI iodide hydroiodic acid

Correct Naming oxoacids Oxoacids contain an oxygen atom in the anion. To name them, again, start with the name of the anion when naming acids. Polyatomic anions are named according to the following rules: • If an element can only form two oxyanions then the one with the least amount of oxygens is given the ending ite . The one containing the most oxygens is given the ending ate . • Elements that can form four polyatomic oxyanions use prefixes in addition to the ite and ate endings. • The oxyanion with the least amount of oxygens is given the prefix hypo and the ending ite (i.e., is hypochlorite). • The oxyanion with the second least amount of oxygens is simply given the ite ending (i.e., is chlorite). • The oxyanion with the second most amount of oxygens is simply given the ate ending (i.e., is chlorate). • Finally, the oxyanion with the most amount of oxygens is given the prefix per and the ending ate (i.e., is perchlorate). When naming the acid of an oxoanion the endings of the anion are modified as follows: • ate changes to ic (retain prefixes) and • ite changes to ous (retain prefixes). The word acid is added following the new name of the anion to complete the name of the acid. Part B What is the formula for chlorous acid? Express your answer as a chemical formula. Hint 1. How to approach the problem Since the prefix hydro is absent in this name, the anion must be a polyatomic ion. The suffix ous tells us that the anion by itself ends with the suffix ite . Hint 2. Identify the anion name What is the name of the anion to which the hydrogen ion is bound in chlorous acid? Spell out the full name of the ion (e.g., nitrate). ANSWER: Hint 3. Identify the anion formula ClO − ClO 2 − ClO 3 − ClO 4 − chlorite

What is the formula for the chlorite ion? ANSWER: ANSWER: Correct Part C What is the name of the acid whose formula is ? Spell out the full name of the acid. Hint 1. Determine the anion formula What is the formula for the anion in ? Express your answer as an ion. ANSWER: Hint 2. Determine the anion name What is the name of the anion, , to which the hydrogen ions are bound? Spell out the full name of the ion (e.g., nitrate). ClO − ClO 2 − ClO 3 − ClO 4 − HClO 2 C H 2 O 3 C H 2 O 3 CO 3 2 − CO 3 2 −

ANSWER: ANSWER: Correct When carbon dioxide dissolves in ocean water, soda, or blood it forms carbonic acid. Part D What is the formula for sulfurous acid? Express your answer as a chemical formula. Hint 1. Determine the anion name What is the name of the anion to which the hydrogen ion is bound in sulfurous acid? Spell out the full name of the ion (e.g., nitrate). ANSWER: Hint 2. Determine the anion formula What is the formula for the sulfite ion? Express your answer as an ion. ANSWER: carbonate carbonic acid sulfite SO 3 2 −

ANSWER: Correct Sulfurous acid forms from sulfur dioxide, contributing to acid rain. Functional Groups Learning Goal: To recognize and identify specific functional groups within organic molecules. Functional groups are groups of atoms that replace atoms on hydrocarbons. Functional groups allow complex organic molecules to be built from simple hydrocarbon starting molecules. Some of the more important functional groups are listed in the table here. Class Functional group Formula Alcohols Amines Aldehydes Ketones Carboxylic acids Part A Classify the organic compounds by the class of their functional group. Drag the appropriate items to their respective bins. S H 2 O 3 H − OH R − OH − NH 2 R − NH 2 O     ||     − C − H O || R − C − H O || − C − O || R − C − R O        ||        − C − OH O     ||     R − C − OH

Hint 1. How to approach the problem Identify the functional group in the organic compound and then find the corresponding class from the table provided in the introduction. Hint 2. How to distinguish among oxygen-containing functional groups Keep in mind that ketone functional groups must be bound to internal carbon atoms, not to a carbon atom at the end of a chain. Also, alcohol groups are usually written with the oxygen next to the carbon to which it binds ( or ), whereas aldehyde groups are written with the hydrogen atom next to the carbon ( ). Hint 3. Identify the functional group in The functional group in is ANSWER: ANSWER: COH HOC CHO C COC H 3 H 3 C COC H 3 H 3 − CHO − COOH − CO −

Correct Organic Functional Groups In chemistry there are several functional groups that are used to describe organic molecules. These functional groups often look similar or have similar names. For example, amines and amides are both organic compounds that contain nitrogen. Amines are ammonia derivatives, in which at least one hydrogen atom is replaced by an organic substituent. Amides contain the carbonyl group ( ) bonded to a nitrogen atom. Below are some additional functional groups that are commonly seen in organic chemistry: Class Functional group Reset Help C=O HOC C C H 2 H 2 H 3 C C C C COOH H 3 H 2 H 2 H 2 N C 6 H 5 H 2 C COC H 3 H 3 C C CHO H 3 H 2

Amine Amide attached to a Aldehyde (carbonyl) and Ketone between carbon atoms Ether between two atoms Ester attached to a Alcohol (hydroxyl) Carboxylic acid attached to a Part A Classify each molecule as an amine, amide, or neither. Drag each item to the appropriate bin. Hint 1. How to approach the problem 1. Look for molecules with bonded to . Place these molecules in the "Amide" bin. 2. Look for molecules with bonded to alkyl groups. Place these molecules in the "Amine" bin. 3. Place the remaining molecules in the "Neither amine nor amide" bin. Hint 2. Identify an amide Which of the following is an amide? ANSWER: | − N − C=O − N − C=O − H C=O − O − C C=O − O − − OH C=O − OH N C=O N

ANSWER: N(CH 3 ) 3 C − NH − C H 3 H 3 C − C H 3 C || O N | H H 3 − N C 6 H 5 H 2 Reset Help

Correct Amines are bases because the lone pair of electrons on the nitrogen atom can act as a acceptor. After accepting the ion, the amine becomes an ammonium salt that is more soluble in water than the original amine. Amides are neutral; they do not act as acids or bases. The carbonyl group prevents the lone pair of electrons on the nitrogen atom from bonding with a hydrogen ion. Part B Classify each molecule as an aldehyde, ketone, or neither. Drag each item to the appropriate bin. Hint 1. Identify suffixes in aldehyde and ketone nomenclature What are the suffixes for aldehydes and ketones? Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer. ANSWER: H + H +

Hint 2. Identify the general structure of an aldehyde Which choice shows the general structure of an aldehyde? ANSWER: Reset Help ol oate ane ene yne oic acid 1. The suffix applied to the name of an aldehyde is al . 2. The suffix applied to the name of a ketone is one .

Hint 3. Identify the general structure of a ketone Which choice shows the general structure of a ketone? ANSWER: O    ∥   R − C − OH O        ∥       R − C − O − R ′ O ∥ R − C − H R − OH O ∥ R − C − R ′

ANSWER: O    ∥   R − C − OH O        ∥       R − C − O − R R − OH O ∥ R − C − H O ∥ R − C − R ′

Correct Part C Ethers and esters are commonly confused. Ethers contain only single bonds and have one oxygen atom connected to two carbon atoms. Esters contain a carboxyl group and have two oxygen atoms. Classify each molecule as an ether, ester, or neither. Drag each item to the appropriate bin. Hint 1. Identify the suffix for an ester ANSWER: Reset Help hexanal pentanal butanal 3-heptanone butanone methyl propanoate propanoic acid methyl hexanoate

Hint 2. Identify the general structure of an ether Which choice shows the general structure of an ether? ANSWER: Hint 3. Identify the general structure of an ester Which choice shows the general structure of an ester? ANSWER: The suffix for an ester is . al ane ol oate oic acid O        ∥       R − C − O − R ′ R − OH O ∥ R − C − R ′ O    ∥   R − C − OH R − O − R

ANSWER: O        ∥       R − C − O − R O ∥ R − C − R ′ O ∥ R − C − H R − OH O    ∥   R − C − OH

Correct Part D Draw an ester with the formula . Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default. Include all hydrogen atoms. Hint 1. Identify the general structure of an ester Which choice shows the general structure of an ester? ANSWER: Reset Help C 3 H 6 O 2 diethyl ether ethyl hexanoate methyl benzoate 3-methyl-2-hexanol cyclohexanone palmitic acid

Hint 2. Identify the number of double bonds How many double bonds can a structure with the formula have? ANSWER: ANSWER: O        ∥       R − C − O − R O    ∥   R − C − OH O ∥ R − C − R ′ O ∥ R − C − H R − OH C 3 H 6 O 2 0 1 2

Correct Simple Organic Compounds Organic molecules generally consist of a chain of carbon atoms surrounded by hydrogen atoms or functional groups. The number of carbon atoms in the main chain is specified by using a prefix, the most common of which are listed in the table.

Prefix Meaning meth 1 eth 2 prop 3 but 4 pent 5 hex 6 hept 7 oct 8 non 9 A carbon atom will form four bonds with other atoms, while oxygen will form two bonds and hydrogen will form one bond. For this problem, draw all hydrogen atoms explicitly. Part A Draw a heptane molecule. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all hydrogen atoms. Hint 1. Determine the number of carbon atoms in heptane How many carbon atoms are in a molecule of heptane? Express your answer numerically as an integer. ANSWER: Hint 2. Determine the number of bonds for each carbon atom How many bonds does a carbon atom form in organic molecules? Express your answer numerically as an integer. ANSWER: 7

Hint 3. Draw propane as an example As an example, draw propane, an alkane with three carbon atoms. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all hydrogen atoms. ANSWER: 4 H − − − − H C | H | H C | H | H C | H | H

ANSWER:                      C C C H H H H H H H H + - C H O N S P F Br Cl I X More

Correct The formula for heptane is . Heptane is classified as an alkane.                      C C C C C C C H H H H H H H H H H H H H H H H + - C H O N S P F Br Cl I X More C 7 H 16

Part B Draw a molecule of 2-pentanol. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all hydrogen atoms. Hint 1. Determine the number of carbon atoms in pentanol How many carbon atoms are in a molecule of pentanol? Express your answer numerically as an integer. ANSWER: Hint 2. Determine the meaning of the ol ending What does the ending ol indicate? ANSWER: Hint 3. Determine the meaning of the number "2" What does the number "2" indicate in 2-pentanol? ANSWER: 5 the presence of a double bond the presence of an group − OH the presence of a ring structure the presence of a triple bond

Hint 4. How to place an group Hydrogen atoms only form one bond whereas oxygen atoms can form two bonds. Therefore, you must attach the oxygen atom to the carbon chain rather than the hydrogen atom. You must also explicitly show the bond between the oxygen atom and the hydrogen atom so that the system will be able to interpret your answer correctly. Hint 5. Classify butanol structures by the position of the group To help you understand how the position of the group relates to the name of the compound, consider these structures of butanol. Match each structure to its name. Drag the appropriate items to their respective bins. ANSWER: There are two groups. − O − H There is an group on the middle carbon atom. − O − H There is an group on the second carbon atom. − O − H There are two carbon atoms. There are two pentanol molecules. − O − H Correct: Incorrect: − C − O − H − C − H − O Correct: Incorrect: − C − O − H − C − OH − O − H − O − H

ANSWER: Reset Help H − − − − − H C | H | O | H C | H | H C | H | H C | H | H H − − − − − H C | H | H C | H | H C | H | H C | H | O | H H − − − − − H C | H | H C | H | O | H C | H | H C | H | H H − − − − − H C | H | H C | H | H C | H | O | H C | H | H

Correct The formula for pentanol is or . Pentanol is classified as an alcohol.                      C C C C C O H H H H H H H H H H H H + - C H O N S P F Br Cl I X More O C 5 H 12 OH C 5 H 11

Part C What can be said about the relationship between the number of hydrogen atoms and the number of carbon atoms in an alkane or alcohol with only single bonds? Hint 1. How to approach the problem Notice in the structures from Parts A and B that the carbon atoms at each end of the chain are bonded to three hydrogen atoms each. However, the inner carbon atoms are either directly or indirectly bonded to two hydrogen atoms each. Think about how you could predict the number of hydrogen atoms for a carbon chain of any length. Each carbon atom is bonded to at least two hydrogen atoms, and each of the end carbon atoms has one extra hydrogen atom. Hint 2. Count the hydrogen atoms in heptane Look at the heptane molecule from Part A. There are seven atoms. How many hydrogen atoms are there? Express your answer numerically as an integer. ANSWER: Hint 3. Count the hydrogen atoms in pentanol Look at the pentanol molecule from Part B. There are five atoms. How many hydrogen atoms are there? Express your answer numerically as an integer. ANSWER: ANSWER: C 16 atoms H C 12 atoms H

Correct For this reason, the general formula for alkanes is expressed as . Similarly, the general formula for alcohols could be expressed as or . Isomers Isomers are molecules with the same chemical formula, but different structures. Part A Which choice is a constitutional isomer of this molecule? Hint 1. How to approach the problem If you were holding a molecular model of this compound in your hand, you would have to pull it apart and reform it in some other arrangement to produce a constitutional isomer. Flipping the molecule over horizontally or vertically would not produce a consitutional isomer. Also, rotation around carbon-carbon single bonds does not result in constitutional isomers. ANSWER: To arrive at the number of hydrogen atoms, triple the number of carbon atoms, then subtract three. there is no consistent formula. triple the number of carbon atoms, then subtract five. triple the number of carbon atoms. double the number of carbon atoms. double the number of carbon atoms, then add two. C n H 2 n +2 O C n H 2 n +2 OH C n H 2 n +1 C − − − C − C H 3 CH | Cl CH | CH 3 H 2 H 3

Correct Part B Which of the following molecules has a conformational isomer? Hint 1. How to approach the problem Conformational isomers have either an eclipsed or staggerred conformation. This requires a carbon-carbon single bond for rotation. Hint 2. Determine the bonding of the compounds The bonds between the carbon atoms are shown for the compounds in the problem. Select the compound in which the bonds are drawn incorrectly. ANSWER: C − C − − − C H 3 H 2 CH | CH 3 CH | Cl H 3 C − − − C − C H 3 CH | Cl CH | CH 3 H 2 H 3 C − − − C − C H 3 CH | CH 3 CH | Cl H 2 H 3 HC ≡ CH C=C H 2 H 2 C − C − C − C H 3 H 2 H 2 H 3 C − C ≡ C − C H 2 H 2

ANSWER: Correct Part C Rank these conformational isomers in order of decreasing potential energy. Rank from hightest to lowest potential energy. To rank items as equivalent, overlap them. Hint 1. Relationship between potential energy and stability The conformational isomer that has the most potential energy will be the least stable isomer. The isomer with the lowest potential energy will be the most stable isomer. Hint 2. Factors that affect potential energy Conformational isomers are the result of rotation about a carbon-carbon single bond. The substituents are shifted in space when this rotation occurs. There are two extreme possibilities of conformations: eclipsed and staggered. Isomers with an eclipsed arrangement have the greatest potential energy because the atoms are positioned to have the maximum overlap with each other when viewed along the axis of the carbon-carbon bond. Isomers with a staggered arrangement have the least potential energy as the atoms are positioned to have the least possible overlap with each other when viewed along the axis of the carbon-carbon bond. ANSWER: CHCH C C H 2 H 2 C C C C H 3 H 2 H 2 H 3 C CCC H 2 H 2

Correct Aromatic Nomenclature Learning Goal: To understand aromatic compound characteristics and nomenclature Aromatic compounds are a class of organic compound characterized by planar carbon-ring structures with alternating single and double bonds. According to Huckel's rule aromatic compounds contain electrons. The most common aromatic ring is benzene with six electrons: Reset Help The correct ranking cannot be determined. 4 n + 2 π π

Aromatic compounds are characterized by increased stability owing to the charge delocalization and resonance. Part A Identify the aromatic compounds. Drag each item to the appropriate bin. Hint 1. How to approach the problem Aromatic compounds must contain alternating single and double bonds, be flat, and obey Huckel's rule. First look for rings containing alternating single and double bonds, then verify that Huckel's rule is followed. Finally, if necessary, determine whether the molecule is flat. Hint 2. Determine whether Huckel's rule is obeyed for a sample compound Huckel's rule states that aromatic compounds contain electrons. For example, an aromatic compound may contain electrons or electrons. Does cyclopentadiene, which contains two double bonds, obey Huckel's rule? ANSWER: ANSWER: 4 n + 2 π 4 × 1 + 2 = 6 π 4 × 2 + 2 = 10 π Yes, cyclopentadiene contains six electrons and, therefore, obeys Huckel's rule. π Yes, cyclopentadiene contains four electrons and, therefore, obeys Huckel's rule. π No, cyclopentadiene contains six electrons and, therefore, does not obey Huckel's rule. π No, cyclopentadiene contains four electrons and, therefore, does not obey Huckel's rule. π

Correct Naming benzene rings Benzene rings with substituents are identified using the name of the substituent and the root name benzene . If there are two substituents, a prefix is used to describe the placement of the substituents. If they are on adjacent carbon atoms, it is termed a 1,2 relationship (they are on C1 and C2), and the prefix ortho- is used (for example, ortho -dichlorobenzene). If the two substituents have a 1,3 relationship, the prefix meta- is used. The prefix para- signifies substituents in a 1,4 relationship. Part B Draw the structure of the aromatic compound methylbenzene (toluene). Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default. Include all hydrogen atoms. Reset Help

Hint 1. How to approach the problem Start by drawing benzene. Then replace one hydrogen atom with a methyl substituent. Hint 2. Draw benzene Draw the structure of benzene. Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default. Include all hydrogen atoms. ANSWER:

Hint 3. Identify the substituent What is the formula for a methyl group? Express your answer as a chemical formula. ANSWER: ANSWER: CH 3

Correct Part C Draw the structure of the aromatic compound para -aminochlorobenzene ( para -chloroaniline). Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default. Include all hydrogen atoms. Hint 1. How to approach the problem Start by drawing benzene. Then, replace one hydrogen atom with a chloro group. Next, identify the position para to the chloro group. Replace the hydrogen atom in the para position with the amino group. Hint 2. Determine the substituents What are the formulas for the chloro and the amino group, respectively? Express your answers as chemical formulas separated by a comma. ANSWER: Hint 3. Identify the para position In which of the following structures are the X atoms located in the para position? ANSWER: , Cl NH 2

ANSWER:

Correct This compound has many common names, for example 4-chloro-1-aminobenzene or p -chloroaniline. Part D Name the following compound:

Spell out the full name of the compound. Hint 1. How to approach the problem Start by identifying the position of the substituents relative to each other as either ortho , meta , or para . When writing the name, first write the position of the substituents, then write each functional group alphabetically. Finally, the compound name will end in benzene. Hint 2. Identify the position of the substituents Sort each of the following based on the position of the X substituents. Drag the appropriate items to their respective bins. ANSWER:

Hint 3. Name the Br substituent When writing the names of compounds, what is the group name for ? For example, the group name for a group on a benzene ring is cyano. Spell out the full name of the group. ANSWER: Hint 4. Name the substituent When writing the names of compounds, what is the group name for ? Reset Help Br CN bromo NO 2 NO 2

For example, the group name for a group on a benzene ring is cyano. Spell out the full name of the group. Hint 1. The name of the ion The ion is called the nitrite ion. To create the group name, drop the suffix from the ion name and add the letter o . ANSWER: ANSWER: Correct Score Summary: Your score on this assignment is 95.9%. You received 8.63 out of a possible total of 9 points. CN NO 2 NO 2 − nitro meta-bromonitrobenzene

3.1 Gases Due: 11:59pm on Sunday, October 1, 2023 You will receive no credit for items you complete after the assignment is due. Grading Policy Boyle's Law In the mid 1600s, the Irish chemist Robert Boyle carried out experiments that determined the quantitative relationship between the pressure and volume of a gas. His data showed that for gas at a constant temperature, pressure and volume are inversely proportional . Part A According to Boyle's law, for a fixed quantity of gas at a given temperature, what quantity relating pressure and volume is constant? Hint 1. Inverse proportionality Two quantities are inversely proportional if one quantity is equal to a constant divided by the other. Thus if one of the quantities is halved, the other will double. Boyle's law states that volume and pressure are inversely proportional. Thus, for example, if , and the volume of a gas at constant temperature is halved, so that , the pressure would have to double, so that . ANSWER: Correct Application of Boyle's law A 12-liter tank contains helium gas pressurized to 160 . P V V P V = 6 P = 5 V = 3 P = 10 P V P × V P + V V P atm

Part B What size tank would be needed to contain this same amount of helium at atmospheric pressure (1 )? Express the size in liters to three significant figures. Hint 1. Inverse proportionality The pressure is being decreased by a factor of 160. Therefore the volume must increase by a factor of 160. ANSWER: Correct Part C How many 3-liter balloons at atmospheric pressure (1 ) could the 12-L helium tank pressurized to 160 fill? Keep in mind that an "exhausted" helium tank is not empty. In other words, once the gas inside the tank reaches atmospheric pressure, it will no longer be able to fill balloons. Express the number of balloons to three significant figures. Hint 1. Determine the volume of "unused" helium You are told you have a 12-L helium tank. What volume of unused helium will remain in the tank once it reaches atmospheric pressure? Express the volume in liters to two significant figures. ANSWER: Hint 2. Determine the volume of "usable" helium You have determined in Part B that you have a total of 1920 liters of uncompressed helium. Subtract the number of liters of unused helium to obtain the number of liters of usable helium. What is the total volume of uncompressed helium that can be used toward filling balloons? Express the volume in liters to four significant figures. atm 1920 L atm atm 12 L

ANSWER: Correct It is best to keep extra digits and round only at the end of your calculation. ANSWER: Correct Interactive Simulation—Avogadro's Law Avogadro's law states that the volume of a gas at a given temperature and pressure is directly proportional to the number of moles of the gas. Click on the image to explore this simulation , which demonstrates the principles of Avogadro's law. In the simulation, you should see a container with a pressure cylinder. You can set the temperature from the dropdown list provided in the simulation. The graph displayed on the right side of the container indicates the relationship between the volume in liters ( ) and the number of moles. The text box provided in the lower right corner of the simulation gives 1908 L 636 balloons L

the value of pressure, , volume, , and number of moles, . Part A In the simulation, set the temperature of the sample gas to 304 . Click on the "Run" button and observe how the number of moles of the gas varies with an increase in the gas’s volume. Then, click on the "Reset" button, set the temperature of the sample gas to 504 , and click on the "Run" button again.Observe the graph produced in the simulation and select the correct options listed below. Check all that apply. Hint 1. What Avogadro's law is Avogadro's law states the relationship between the volume and the number of moles of a gas at constant temperature and pressure. It states that at constant temperature and pressure, the volume of gas is directly proportional to the number of moles of the gas: In other words, at constant temperature and pressure, if you increase the number of moles of a gas, its volume will also increase. Thus, the ratio between the volume and the number of moles of gas is constant: Hint 2. Calculate the volume of the gas when the number of moles of the gas is doubled In the simulation, set the temperature, , to 304 and run the simulation. Click on the "Stop" button to stop the simulation when the number of moles of the gas, , reaches 0.40 . Note the corresponding volume ( ) of the gas from the text box provided in the lower right corner of the simulation. If the number of moles of gas exceeds 0.40 , observe the reading for the volume corresponding to 0.4 from the given graph. Then, click the "Run" button again and note the volume ( ) of the gas when the number of moles of the gas, , reaches 0.80 . If the number of moles of gas exceeds 0.80 , observe the reading for the volume corresponding to 0.80 from the given graph.Select the answer that correctly states the volume at each of the given numbers of moles of the gas. ANSWER: ANSWER: p V n K K V   ∝   n V / n   =   constant T K n mol V mol mol V n mol mol mol 20.0 , 10.0 L L 24.9 , 24.9 L L 10.0 , 20.0 L L 0.40 , 0.80 mol mol

3.2 Gases 2 Due: 11:59pm on Sunday, October 22, 2023 You will receive no credit for items you complete after the assignment is due. Grading Policy ± Effusion and Molar Mass The rate of effusion of a gas, , is inversely proportional to the square root of its molar mass, . The relative rate of two different gases is expressed as where and are the effusion rates of two gases and and are their respective molar masses. Part A In an effusion experiment, it was determined that nitrogen gas, , effused at a rate 1.812 times faster than an unknown gas. What is the molar mass of the unknown gas? Express your answer to four significant figures and include the appropriate units. Hint 1. How to approach the problem If we designate nitrogen as gas #1 and the unknown as gas #2, then Graham's law can be rewritten as Even though we do not know the value of , we do know that it effuses 1.812 times faster than the unknown, allowing us to substitute the value 1.182 for the quantity Thus The final step is to plug in the molar mass of and solve for . Hint 2. Rearrange Graham's law If represents the ratio of effusion rates, r M = r 1 r 2 M 2 M 1 −− − √ r 1 r 2 M 1 M 2 N 2 = r N 2 r unknown M unknown M N 2 − − −−−− √ r N 2 / . r N 2 r unknown 1.812 = M unknown M N 2 − − −−−− √ N 2 M unknown x x = M 2 M 1 −− − √

what is the equation for ? Express in terms of and . ANSWER: Correct Hint 3. Determine the molar mass of nitrogen gas What is the value of , the molar mass of nitrogen gas? Express your answer to four significant figures and include the appropriate units. ANSWER: Correct ANSWER: Correct Effusion can be used to separate a mixture of gases provided that their molar masses differ by a significant amount. Multiple-step effusions can be used to separate gases with similar molar masses. Part B Elemental analysis of the unknown gas from Part A revealed that it is 30.45 and 69.55 by mass. What is the molecular formula for this gas? Express your answer as a chemical formula. Hint 1. Determine the empirical formula M 2 M 2 x M 1 = M 2 x 2 M 1 M N 2 = 28.01 M N 2 g mol = 91.98 M g mol % N % O

Based on the elemental analysis, what is the empirical formula for the unknown gas? Express your answer as a chemical formula. Hint 1. Determine the number of moles of nitrogen in the sample At 30.45 a 100- sample of this gas would contain 30.45 of . Convert this value to moles. Express your answer to four significant figures and include the appropriate units. Hint 1. Identify how to convert mass to moles If is the mass (in ) and is the molar mass (in ) , what operation results in units of moles? ANSWER: ANSWER: Correct Compare this number of moles to the number of moles of oxygen atoms in the sample to determine the empirical formula. Hint 2. Determine the number of moles of oxygen in the sample At 69.55 a 100- sample of this gas would contain 69.55 of . Convert this value to moles. Express your answer to four significant figures and include the appropriate units. Hint 1. Identify how to convert mass to moles If is the mass (in ) and is the molar mass (in ) , what operation results in units of moles? ANSWER: %N g g N x g y g/mol x × y x / y y / x 2.174 mol %O g g O x g y g/mol

ANSWER: Correct Compare this number of moles to the number of moles of nitrogen atoms in the sample to determine the empirical formula. Hint 3. How to determine the mole ratio Now that you know the relative numbers of moles of and you need to express them as a whole-number ratio. One strategy for doing so is to divide each value by the lower of the two values: The resulting integers should be used as the subscripts in the formula . ANSWER: Correct Now consider the molar mass from Part A. Adjust the subscripts of the empirical formula as necessary to arrive at a molar mass of about 92 , while keeping the 1:2 ratio of nitrogen atoms to oxygen atoms. ANSWER: y / x x / y x × y 4.347 mol N O = x 2.174   mol   N 2.174 = y 4.327   mol   O 2.174 N x O y NO 2 g/mol N 2 O 4

Correct ± The van der Waals Equation Learning Goal: To be able to predict and calculate properties of real gases. Kinetic molecular theory makes certain assumptions about gases that are, in fact, not true for real gases. Therefore, the measured properties of a gas are often slightly different from the values predicted by the ideal gas law. The van der Waals equation is a more exact way of calculating properties of real gases. The formula includes two constants, and , that are unique for each gas. The van der Waals equation is , where is the pressure, the number of moles of gas, the volume, the temperature, and the gas constant. All real gases possess intermolecular forces that can slightly decrease the observed pressure. In the van der Waals equation, the variable is adjusted to , where is the attractive force between molecules. The volume of a gas is defined as the space in which the gas molecules can move. When using the ideal gas law we make the assumption that this space is equal to the volume of the container. However, the gas molecules themselves take up some space in the container. So in the van der Waals equation, the variable is adjusted to be the volume of the container minus the space taken up by the molecules, , where is the volume of a mole of molecules. Part A In general, which of the following gases would you expect to behave the most ideally even under extreme conditions? Hint 1. How to approach the problem An ideal gas molecule takes up no space and has no attraction to other molecules. Therefore, choose the substance with the smallest molecules and the least amount of intermolecular attraction. Hint 2. Relative sizes C, O, and N are close together in the periodic table. Therefore, and are about the same size. Hydrogen is in the row above C, O, and N, and therefore is the smallest of the three molecules. Hint 3. Relative intermolecular attraction is a polar molecule and therefore carbon monoxide gas is held together by dipole attraction. In contrast, and are both nonpolar molecules, and therefore a b ( P + ) ( V − nb ) = nRT an 2 V 2 P n V T R P P + an 2 V 2 a V V − nb b CO N 2 H 2 CO N 2 H 2

nitrogen gas and hydrogen gas are both held together by dispersion forces. In general, dispersion is the weakest intermolecular force. ANSWER: Correct Part B In general, which of the following gases would you expect to behave the least ideally under extreme conditions? Hint 1. How to approach the problem An ideal gas molecule takes up no space and has no attraction to other molecules. Therefore, choose the substance with the largest molecules and the greatest amount of intermolecular attraction. Hint 2. Relative sizes C, O, and N are close together in the periodic table. Therefore, and are about the same size. Hydrogen is in the row above C, O, and N, and therefore is the smallest of the three molecules. Hint 3. Relative intermolecular attraction Carbon monoxide, , is a polar molecule, and therefore carbon monoxide gas is held together by dipole attraction. In contrast, and are both nonpolar molecules, and therefore nitrogen gas and hydrogen gas are both held together by dispersion forces. In general, dispersion is the weakest intermolecular force ANSWER: N 2 CO H 2 CO N 2 H 2 CO N 2 H 2

Correct Part C 13.0 moles of gas are in a 8.00 tank at 20.5 . Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are and . Express your answer with the appropriate units. Hint 1. How to approach the problem First, use the ideal gas law to solve for . Then use the same values in the van der Waals equation and solve for . Subtract the two values to obtain the difference in pressure. Hint 2. Find the pressure of an ideal gas Using the ideal gas equation, , calculate the pressure of the ideal gas. Express your answer with the appropriate units. ANSWER: Hint 3. Determine the pressure of methane Using algebra, we can solve the van der Waals equation for : Using the rearranged van der Waals equation, calculate the pressure of methane. Express your answer with the appropriate units. ANSWER: N 2 CO H 2 L C ∘ a = 2.300 ⋅ L 2 atm/mol 2 b = 0.0430   L/mol P P P PV = nRT 39.2 atm P P = − nRT V − bn an 2 V 2

ANSWER: Correct The answer to this question represents the error incurred by using the ideal gas law for real gases. This error is very tiny at low pressure and high temperature, but becomes quite significant for compressed gases (high pressure). ± Ideal versus Real Gases The ideal gas law describes the relationship among the volume of an ideal gas ( ), its pressure ( ), its absolute temperature ( ), and number of moles ( ): Under standard conditions, the ideal gas law does a good job of approximating these properties for any gas. However, the ideal gas law does not account for all the properties of real gases such as intermolecular attraction and molecular volume, which become more pronounced at low temperatures and high pressures. The v an der Waals equation corrects for these factors with the constants and , which are unique to each substance: The gas constant is equal to 0.08206 . Part A A 3.00- flask is filled with gaseous ammonia, . The gas pressure measured at 22.0 is 1.25 . Assuming ideal gas behavior, how many grams of ammonia are in the flask? Express your answer to three significant figures and include the appropriate units. 36.0 atm pressure difference = -3.13 atm V P T n PV = nRT a b ( P + ) ( V − nb ) = nRT an 2 V 2 R L ⋅ atm/(K ⋅ mol) L NH 3 C ∘ atm

Hint 1. How to approach the problem Gas law calculations require the use of absolute temperature, so start by converting the temperature from the Celsius scale to the Kelvin scale. Next, use the ideal gas law to calculate the number of moles. Finally, convert moles to grams using the molar mass of ammonia. Hint 2. Convert temperature to kelvins Convert 22.0 to the Kelvin scale. Express your answer to three significant figures and include the appropriate units. ANSWER: Hint 3. Calculate the number of moles You can rearrange the ideal gas law ( ) given in the introduction to solve for moles, . You will also need to convert the temperature to kelvins. How many moles of ammonia are present in the flask? Express your answer to three significant figures and include the appropriate units. ANSWER: Hint 4. Calculate the molar mass of ammonia What is the molar mass of ammonia, ? Express your answer to three significant figures and include the appropriate units. ANSWER: ANSWER: C ∘ = 295 T K PV = nRT n 0.155 mol NH 3 molar mass of = 17.0 NH 3 g mol mass of = 2.63 NH 3 g

Correct An ideal gas concept assumes that the volume of gas particles is negligible compared to the total volume the gas occupies and there are no attractive forces between individual gas particles. All real gases deviate slightly from the ideal behavior. Under most conditions the difference is very small and we can still use the ideal gas law. But at higher pressures the deviation from the ideal behavior becomes greater, and we need to take both the size of individual gas particles and the attractive forces between individual gas particles into account. Ideal versus real behavior for gases In the following part you can see how the behavior of real gases deviates from the ideal behavior. You will calculate the pressure values for a gas using the ideal gas law and also the van der Waals equation. Take note of how they differ. Part B If 1.00 of argon is placed in a 0.500- container at 26.0 , what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)? For argon, and . Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem You will calculate the pressure twice. First, use the ideal gas law, and then use the van der Waals equation. Finally, subtract the smaller pressure from the larger pressure to find the difference. Be sure to use the the Kelvin temperature scale for both pressure calculations. Hint 2. Convert temperature to kelvins Convert 26.0 to the Kelvin scale. Express your answer to three significant figures and include the appropriate units. ANSWER: mol L C ∘ a = 1.345 ( ⋅ atm)/mo L 2 l 2 b = 0.03219 L/mol C ∘ = 299 T K

Hint 3. Calculate the pressure using the ideal gas law Calculate the pressure in atmospheres of 1.00 of argon in a 0.500- container at 26.0 . Express your answer to three significant figures and include the appropriate units. ANSWER: Hint 4. Calculate the pressure using the van der Waals equation Using the van der Waals equation, calculate the pressure in atmospheres of 1.00 of argon in a 0.500- container at 26.0 . For argon, and . Express your answer to three significant figures and include the appropriate units. Hint 1. How to rearrange the equation to solve for pressure Use the van der Waals equation, and rearrange it to solve for : Hint 2. Calculate the first term of the rearranged equation Determine the value of Express your answer to three significant figures and include the appropriate units. ANSWER: Hint 3. Calculate the second term of the rearranged equation Determine the value of Express your answer to three significant figures and include the appropriate units. mol L C ∘ 49.1 atm mol L C ∘ a = 1.345 ( ⋅ atm)/mo L 2 l 2 b = 0.03219 L/mol ( P + ) ( V − nb ) = nRT an 2 V 2 P P = − nRT V − nb an 2 V 2 nRT V − nb 52.4 atm an 2 V 2

ANSWER: ANSWER: ANSWER: Correct At high pressures and low temperatures the assumptions of the ideal gas law, which postulate that gas particles are small particles that act independently of one another, begin to break down. Part C In Part B the given conditions were 1.00 of argon in a 0.500- container at 26.0 . You identified that the ideal pressure ( ) was 49.1 , and the real pressure ( ) was 47.1 under these conditions. Complete the sentences to analyze this difference. Match the words and compounds in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer. Hint 1. How to calculate percent difference Percent difference can be calculated by the following equation Since the value should be positive, it is the absolute value of the difference between two values divided by their average. The ratio is multiplied by 100 for the percent. A percent difference is used to compare two values when an exact measurement is unknown. Percent error would be used if an exact value was known. Hint 2. Identify how volume impacts pressure When the volume of the container increases from 0.500 to 2.00 while the number of moles of gas and temperature remain constant, what happens to the pressure within the container? 5.38 atm 47.1 atm = 2.0 − P ideal P real atm mol L C ∘ P ideal atm P real atm percent   difference = | | × 100 value   1 − value   2 (value   1+value   2)/2 L L

ANSWER: Correct As volume increases, pressure decreases. ANSWER: The pressure in the container can be predicted to decrease. The change in pressure in the container cannot be predicted. The pressure in the container can be predicted to increase. Reset Help kinetic energy of the 4.1 molecular volumes of the increase 4.3 The percent difference between the ideal and real gas is 4.2 . This difference is considered significant, and is best explained due to the intermolecular forces between argon atoms. If the volume of the container were increased to 2.00 , you would expect the percent difference between the ideal and real gas to decrease . % % % L

Correct At high pressures and low temperatures, gas molecules are much closer together and so molecular volume and attractive forces between molecules are more pronounced. The result is a decrease in the actual pressure of most real gases when compared with ideal gases. Score Summary: Your score on this assignment is 98.2%. You received 2.95 out of a possible total of 3 points.

4.1 Thermochemistry 1 Due: 11:59pm on Sunday, October 22, 2023 You will receive no credit for items you complete after the assignment is due. Grading Policy Interactive Activity—Applications of Hess’s Law Open the activity depicting Hess's Law . Part A Click the button within the activity, and analyze the relationship between the two reactions that are displayed. The reaction that was on the screen when you started and its derivative demonstrate that the change in enthalpy for a reaction, , is an extensive property. Using this property, calculate the change in enthalpy for Reaction 2. Reaction 1: Reaction 2: Express your answer to four significant figures and include the appropriate units. Hint 1. What an extensive property is The extensive properties are the ones that depend on the amount of substance present in the system. The enthalpy change of a reaction ( ) is an extensive property. If you add or remove some reactants in the reaction, the value of the enthalpy change of the reaction ( ) changes accordingly. For example, consider the formation of water. The reaction can be represented as Δ H (g) + 5 (g) → 3C (g) + 4 O(g), C 3 H 8 O 2 O 2 H 2 Δ = − 2043   kJ H 1 4 (g) + 20 (g) → 12C (g) + 16 O(g), C 3 H 8 O 2 O 2 H 2 Δ =? H 2 Δ H Δ H

, Consider this reaction as Reaction 1 and the enthalpy change of the reaction as . Reaction 1: , . If you reduce the amount of hydrogen gas to half the number of moles, the balanced chemical reaction will be Consider this reaction as Reaction 2. As the amount of reactants is reduced to half, the enthalpy change of the reaction will also be reduced to half. Thus, the enthalpy for Reaction 2 will now be Reaction 2: , . Hint 2. How to approach the problem To determine the reaction enthalpy change for Reaction 2 using Reaction 1, you need to compare the coefficient of reactants in Reaction 1 and Reaction 2, as they represent the number of moles undergoing the reaction. By comparing the two coefficients of the same reactants in Reaction 1 and Reaction 2, find the factor that relates the two reactions. If the coefficient of a reactant in Reaction 2 is changed by factor , then the value of the change in reaction enthalpy for Reaction 2 will also change by factor : and For example, consider the two reactions Reaction 1: , and Reaction 2: , Then so Hint 3. Identify the factor that relates the two reactions 2 + → 2 O H 2 O 2 H 2 Δ H = − 285.83   kJ Δ H 1 2 + → 2 O H 2 O 2 H 2 Δ = − 285.83   kJ H 1 + → O H 2 1 2 O 2 H 2 Δ = = = − 142.92   kJ H 2 Δ H 1 2 − 285.83   kJ 2 + → O H 2 1 2 O 2 H 2 Δ = − 142.92   kJ H 2 x x factor = coefficient   of   a   reactant   in   Reaction   2 coefficient   of   the   same   reactant   in   Reaction   1 Δ = factor × Δ H 2 H 1 2 + → 2 O H 2 O 2 H 2 Δ = − 285.83   kJ H 1 + → O H 2 1 2 O 2 H 2 Δ =? H 2 factor = = coefficients   of     in   Reaction   2 H 2 coefficients   of     in   Reaction   1 H 2 1 2 Δ = factor × Δ = × − 285.83   kJ = − 142.92   kJ H 2 H 1 1 2

Consider the two reactions Reaction 1: and Reaction 2: . Using the coefficients of the reactants, find the factor that relates the two reactions. Which of the following factors relates Reaction 2 to Reaction 1? ANSWER: ANSWER: Correct The enthalpy change of a reaction is an extensive property. When the coefficient before the reactants increases from 1 to 4, the enthalpy change of the reaction also increases 4 times. Thus, the enthalpy change for Reaction 2 would be . Part B Click on the button within the activity and analyze the relationship between the two reactions that are displayed. The reaction that was on the screen when you started and its derivative demonstrate that the change in reaction enthalpy, , changes sign when a process is reversed. Consider the reaction What will be for the reaction if it is reversed? Express your answer with appropriate units. (g) + 5 (g) → 3C (g) + 4 O(g) C 3 H 8 O 2 O 2 H 2 4 (g) + 20 (g) → 12C (g) + 16 O(g) C 3 H 8 O 2 O 2 H 2 5 1 4 4 = -8172 Δ H 2 kJ 4 × ( − 2043)   kJ = − 8172   kJ Δ H B(s) + (g) → (s), 3 2 O 2 B 2 O 3 Δ H = − 1270kJ Δ H

Hint 1. How to approach the problem To determine the change in reaction enthalpy for the reverse of a given reaction, first identify the sign of the reaction enthalpy change for the given reaction. Recall that when reactions are reversed, the sign of the reaction enthalpy is reversed, but its value remains the same. For example, consider the reaction , If the reaction is reversed, the new reaction and its enthalpy will be You can see that the value of enthalpy (285.83 ) remains the same, and only its sign is changed from negative ( ) to positive (+). However, we usually do not indicate the positive sign. Hint 2. Identify the sign of the change in reaction enthalpy for a given reaction What is the sign of the change in reaction enthalpy for the given reaction, ? ANSWER: Correct The sign of the change in reaction enthalpy will be reversed when the reaction is reversed. Hint 3. Identify the sign of the change in reaction enthalpy for a reverse reaction What would be the sign of the change in reaction enthalpy for the reverse of the reaction ? ANSWER: 2 + → 2 O H 2 O 2 H 2 Δ H = − 285.83   kJ 2 O → 2 + ,   H 2 H 2 O 2 Δ H = 285.83   kJ kJ − B(s) + (g) → (s) 3 2 O 2 B 2 O 3 negative positive B(s) + (g) → (s) 3 2 O 2 B 2 O 3 The reverse of the reaction has a positive sign. negative sign.

Correct The reverse reaction is . ANSWER: Correct If the reaction , = -1270 is reversed you get and thus the sign of the change in reaction enthalpy will also be reversed. The reaction enthalpy change for the reverse reaction is . Hess’s law of constant heat summation Hess’s law states that if a reaction is carried out in a series of steps, for the overall reaction will equal the sum of the enthalpy changes for the individual steps. By using Hess’s law, you can calculate the enthalpy change of the reaction. For example, you can calculate the reaction enthalpy change for the combustion of solid sulfur, , to form gas: The combustion of solid sulfur to form gas occurs in the two steps. Step 1: , . Step 2: , . The overall reaction will be obtained by adding these two steps. When you add the two steps, the reaction enthalpy change values for the individual steps should also be added. Remember that when you add the two steps, the molecule that appears on both sides of a reaction arrow is cancelled from both sides of the equation. Thus, by adding step 1 and step 2, you can obtain the reaction enthalpy change for the overall reaction: (s) → B(s) + (g) B 2 O 3 3 2 O 2 1270 kJ B(s) + (g) → (s) 3 2 O 2 B 2 O 3 Δ H kJ (s) → B(s) + (g) B 2 O 3 3 2 O 2 1270kJ Δ H S(s) SO 3 S(s) + (g) → S (g) 3 2 O 2 O 3 SO 3 S(s) + (g) → S (g) O 2 O 2 Δ H = − 297   kJ S (g) + (g) → S (g) O 2 1 2 O 2 O 3 Δ H = − 99   kJ [Math Processing Error]

Hence, the reaction enthalpy change for the overall reaction is . Part C Click on the button within the activity and use the example shown to calculate the reaction enthalpy, , for the following reaction: Use the series of reactions that follow: 1. , . 2. , . 3. , . 4. , . Express your answer with appropriate units. Hint 1. How to approach the problem Rearrange the given series of reactions such that when you add them you will get Then, perform similar operations on the respective reaction enthalpy change values. For example, if you have multiplied one of the reactant coefficients by 4 then the value for that reaction should also be multiplied by 4. Finally, add up the rearranged reactions and the modified changes in enthalpies to get the enthalpy change of the given reaction. Hint 2. The change in enthalpy associated with the operation performed on a reaction When the reaction parameters are altered by performing certain operations, the associated reaction enthalpy changes accordingly. This is because enthalpy change is an extensive property. 1. When you reverse the reactions the value of changes in sign. For example, take the reaction with . When the reaction is reversed, , will be . 2. When you multiply the reactions by any common factor, the reaction enthalpy change is multiplied by the same factor. For example, take the reaction with . If the reactants are times the given reaction, then, and will be 3. When you add the two reactions, you must add the corresponding enthalpy change values. If a common molecule is present on the reactant side of one reaction and on the product side of the other, it is cancelled from both sides. For example, consider the two reactions , , and , . Adding the two reactions, you get − 297   kJ + ( − 99   kJ) = − 396   kJ Δ H C (g) + 2 (g) → C (g) + 2 O(l) H 4 O 2 O 2 H 2 C(s) + 2 (g) → C (g) H 2 H 4 Δ H = − 74.8 kJ C(s) + (g) → C (g) O 2 O 2 Δ H = − 393.5 kJ 2 (g) + (g) → 2 O(g) H 2 O 2 H 2 Δ H = − 484.0 kJ O(l) → O(g) H 2 H 2 Δ H = 44.0 kJ C (g) + 2 (g) → C (g) + 2 O(l) H 4 O 2 O 2 H 2 Δ H Δ H A → B Δ H = x   kJ B → A Δ H − x   kJ A → B Δ H = x   kJ a a A → a B Δ H a × x   kJ A + B → C Δ H = x   kJ A + C → D Δ H = y   kJ [Math Processing Error]

Hint 3. Identify the operations performed to obtain the given reaction Which operations should you perform on the given four reactions to obtain the following reaction: Series of reactions: 1. 2. 3. 4. Check all that apply. ANSWER: ANSWER: C (g) + 2 (g) → C (g) + 2 O(l)? H 4 O 2 O 2 H 2 C(s) + 2 (g) → C (g) H 2 H 4 C(s) + (g) → C (g) O 2 O 2 2 (g) + (g) → 2 O(g) H 2 O 2 H 2 O(l) → O(g) H 2 H 2 Reverse Reaction 4: O(g) → O(l) H 2 H 2 Multiply Reaction 3 by 1/2: (g) + (g) → O(g) H 2 1 2 O 2 H 2 Multiply Reaction 1 by 2: 2C(s) + 4 (g) → 2C (g) H 2 H 4 Reverse Reaction 1: C (g) → C(s) + 2 (g) H 4 H 2 Reverse Reaction 2: C (g) → C(s) + (g) O 2 O 2 Multiply Reaction 4 by 2: 2 O(l) → 2 O(g) H 2 H 2 -890.7 kJ

Correct You need to first reverse Reaction 1, which will reverse the sign of the enthalpy change value. On reversing Reaction 1 you will get You will keep the second reaction as it is, which will allow you to add one of the two oxygen molecules on the reactant side and cancel out the solid carbon from the product side of the reversed Reaction 1 and the reactant side of Reaction 2. By adding Reaction 3, you will have your second oxygen molecule and cancel out the two moles of hydrogen from the product side of the first reaction: Now you need two moles of liquid water on the product side. So, you will now reverse Reaction 4 and multiply it by 2: Adding this equation to the one derived above you get Heat Capacity Learning Goal: To understand the concepts of heat capacity, specific heat, and molar heat capacity. Heat capacity, , is the amount of energy required to raise the temperature of a substance by exactly 1 degree Celsius. The energy needed to warm an object increases as the mass of that object increases. We see this in our everyday life. For example, we know that it takes much more energy to heat a large tank of water than a small cup. Because of this dependence on mass, experimentally determined heat capacities are always reported in terms of the amount of the substance that is heated. One method is to report how much energy it takes to raise the temperature of one mole of a substance by exactly 1 degree Celsius. This value is the molar heat capacity , which has the symbol .The molar heat capacity is given in the units . A second method is to report how much energy it takes to raise the temperature of one gram of a substance by exactly 1 degree Celsius. This value is the specific heat , which has been given the symbol . The units for specific heat are . The heat capacity of a substance is therefore related to the energy needed to raise its temperature by an amount . That is, , where denotes the number of moles of the substance, or , where denotes the number of grams of the substance. Parts A and B The next two questions pertain to silver. Part A The molar heat capacity of silver is 25.35 . How much energy would it take to raise the temperature of 8.80 of silver by 11.8 ? C (g) → C(s) + 2 (g),  Δ H = 74.8   kJ H 4 H 2 [Math Processing Error] 2 O(g) → 2 O(l),  Δ H = 2 × ( − 44.0)   kJ H 2 H 2 [Math Processing Error] C C p J/(mol ⋅ C) ∘ C s J/(g ⋅ C) ∘ q Δ T q = n Δ T C p n q = m Δ T C s m J/mol ⋅ C ∘ g C ∘

Express your answer with the appropriate units. Hint 1. How to approach the problem Be sure to choose the appropriate formula to answer the question. Since you are given the molar heat capacity of silver, you will need to calculate how many moles of silver are being heated. Next, plug in the known variables and solve for the unknown value, in this case the energy needed, . Hint 2. Choosing the correct formula You have been given the molar heat capacity of silver, but the amount of silver has been given in grams, not moles. You could convert the molar heat capacity to a specific heat, but it is easier to convert the mass of silver to moles. Since it is easiest to use the given molar heat capacity, what is the best formula to use in this problem? ANSWER: Hint 3. Calculate the number of moles of silver To use the formula for molar heat capacity, you need to know the number of moles of silver (atomic mass = 108 ) being heated. How many moles of silver are being heated in this problem? Express your answer with the appropriate units. ANSWER: ANSWER: Correct Part B q q = n Δ T C p q = m Δ T C s g/mol 8.15 × 10 − 2 mol = 24.4 q J

What is the specific heat of silver? Express your answer with the appropriate units. Hint 1. How to approach the problem You have been given the molar heat capacity of silver in Part B. The only difference between the units of specific heat and the units of molar heat capacity is that one is in terms of grams and the other is in terms of moles. Therefore, use a conversion factor between grams and moles for silver. Hint 2. What is the molar mass of silver? What is the molar mass of silver? Express your answer with the appropriate units. ANSWER: Hint 3. Find the correct unit factor setup Which unit factor setup is correct? ANSWER: ANSWER: Correct 108 g mol × 25.35   J 1   mol C ⋅ ∘ 1   mol 108   g × 25.35 J 1   mol C ⋅ ∘ 108   g 1   mol 0.235 J ⋅ (g ⋅ C) ∘ − 1

Parts C and D The next two questions pertain to an unknown metal. Part C It takes 46.2 to raise the temperature of an 8.80 piece of unknown metal from 13.0 to 24.7 . What is the specific heat for the metal? Express your answer with the appropriate units. Hint 1. How to approach the problem You are given the values of and . You can calculate by subtracting the two temperatures. Then you can plug all of these values into the formula to solve for . Hint 2. Identify the correct formula In this problem, you have been given the mass, temperature change, and energy added to the system. Which formula correctly rearranges to solve for the specific heat? ANSWER: Hint 3. Calculate the change in temperature The change in temperature, , is given by . What is the value of for the piece of metal described in this part? Express your answer with the appropriate units. ANSWER: J g C ∘ C ∘ q m Δ T C s q = m Δ T C s = C s q m Δ T qm Δ T 1 qm Δ T m Δ T q Δ T − T final T initial Δ T = 11.7 Δ T C ∘

ANSWER: Correct Part D You know that the unknown metal corresponds to a set list of metals. Given the specific heat ( ) calculated in Part C, which of the following could be the metal? ANSWER: Correct Given the list, the specific heat for iron is the closest to that calculated in Part C. However, many substances have similar specific heat values. For example, lead also has a specific heat value around 0.128 . Thus, to differentiate gold and lead other tests would be needed. Specific Heat The heat capacity of an object indicates how much energy that object can absorb for a given increase in that object's temperature. In a system in which two objects of different temperatures come into contact with one another, the warmer object will cool and the cooler object will warm up until the system is at a single equilibrium temperature. Note the difference between the terms molar heat capacity , which has units of , and specific heat , which has units of . = 0.449 C s J ⋅ (g ⋅ C) ∘ − 1 C s iron: = 0.449   J/(g C) C s ⋅ ∘ aluminum: = 0.903   J/(g C) C s ⋅ ∘ titanium: = 0.523   J/(g C) C s ⋅ ∘ gold: = 0.128   J/(g C) C s ⋅ ∘ copper: = 0.385   J/(g C) C s ⋅ ∘ tin: = 0.213   J/(g C) C s ⋅ ∘ J/(g ⋅ C) ∘ J/(mol ⋅ C) ∘ J/(g ⋅ C) ∘