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(a)
Interpretation: To calculate the molar mass of the given substance.
Concept Introduction: Molar mass is defined as the smallest unit of the compound. It is the
(a)
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Answer to Problem 115A
158 g/mol.
Explanation of Solution
The given molecular compound is
There is only one calcium atom in the given compound. The molar mass of calcium atom is 40 g/mol.
There are six hydrogen atoms in the given compound. The molar mass of one hydrogen atom is 1 g/mol. Therefore, the molar mass of six hydrogen atoms will be
There are four carbon atoms in the given compound. The molar mass of one carbon atom is 12 g/mol. Therefore, the molar mass of four carbon atoms will be
There are four oxygen atoms in the given compound. The molar mass of one oxygen atom is 16 g/mol. Therefore, the molar mass of four oxygen atoms will be
Therefore, the molar mass of the compound will be 158 g/mol.
(b)
Interpretation: To calculate the molar mass of the given substance.
Concept Introduction: Molar mass is defined as the smallest unit of the compound. It is the
(b)
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Answer to Problem 115A
98 g/mol
Explanation of Solution
There are three hydrogen atoms in the given compound. The molar mass of one hydrogen atom is 1 g/mol. Therefore, the molar mass of three hydrogen atoms will be
There is only one phosphorous atom in the given compound. The molar mass of Phosphorous atom is 31 g/mol.
There are four oxygen atoms in the given compound. The molar mass of one oxygen atom is 16 g/mol. Therefore, the molar mass of four oxygen atoms will be
The overall molar mass of the compound will be 98 g/mol
(c)
Interpretation: To calculate the molar mass of the given substance.
Concept Introduction: Molar mass is defined as the smallest unit of the compound. It is the
(c)
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Answer to Problem 115A
342 g/mol.
Explanation of Solution
There are twelve carbon atoms in the given compound. The molar mass of one carbon atom is 12 g/mol. Therefore, the molar mass of twelve carbon atoms will be
There are 22 hydrogen atoms in the given compound. The molar mass of one hydrogen atom is 1 g/mol. Therefore, the molar mass of 22 hydrogen atoms will be
There are eleven oxygen atoms in the given compound. The molar mass of one oxygen atom is 16 g/mol. Therefore, the molar mass of eleven oxygen atoms will be
The overall molar mass of the compound will be 342 g/mol.
(d)
Interpretation: To calculate the molar mass of the given substance.
Concept Introduction: Molar mass is defined as the smallest unit of the compound. It is the
(d)
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Answer to Problem 115A
331 g/mol
Explanation of Solution
There is one lead atom in the given compound. The molar mass of the lead atom is 207 g/mol.
There are two nitrogen atoms in the given compound. The molar mass of one nitrogen atom is 14 g/mol. Therefore, the molar mass of the two nitrogen atoms will be
There are six oxygen atoms in the given compound. The molar mass of one oxygen atom is 16 g/mol. Therefore, the molar mass of six oxygen atoms will be
The overall molar mass of the given compound will be 331 g/mol
Chapter 14 Solutions
Chemistry 2012 Student Edition (hard Cover) Grade 11
- Can you please help mne with this problem. Im a visual person, so can you redraw it, potentislly color code and then as well explain it. I know im given CO2 use that to explain to me, as well as maybe give me a second example just to clarify even more with drawings (visuals) and explanations.arrow_forwardPart 1. Aqueous 0.010M AgNO 3 is slowly added to a 50-ml solution containing both carbonate [co32-] = 0.105 M and sulfate [soy] = 0.164 M anions. Given the ksp of Ag2CO3 and Ag₂ soy below. Answer the ff: Ag₂ CO3 = 2 Ag+ caq) + co} (aq) ksp = 8.10 × 10-12 Ag₂SO4 = 2Ag+(aq) + soy² (aq) ksp = 1.20 × 10-5 a) which salt will precipitate first? (b) What % of the first anion precipitated will remain in the solution. by the time the second anion starts to precipitate? (c) What is the effect of low pH (more acidic) condition on the separate of the carbonate and sulfate anions via silver precipitation? What is the effect of high pH (more basic)? Provide appropriate explanation per answerarrow_forwardPart 4. Butanoic acid (ka= 1.52× 10-5) has a partition coefficient of 3.0 (favors benzene) when distributed bet. water and benzene. What is the formal concentration of butanoic acid in each phase when 0.10M aqueous butanoic acid is extracted w❘ 25 mL of benzene 100 mL of a) at pit 5.00 b) at pH 9.00arrow_forward
- Calculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 Group of answer choices 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 choices: 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0arrow_forward
- Please solvearrow_forwardRank the compounds in each group below according to their reactivity toward electrophilic aromatic substitution (most reactive = 1; least reactive = 3). Place the number corresponding to the compounds' relative reactivity in the blank below the compound. a. CH₂F CH3 F b. At what position, and on what ring, is bromination of phenyl benzoate expected to occur? Explain your answer. :0: C-O phenyl benzoate 6.Consider the reaction below to answer the following questions. A B C NO₂ FeBr3 + Br₂ D a. The nucleophile in the reaction is: BODADES b. The Lewis acid catalyst in the reaction is: C. This reaction proceeds d. Draw the structure of product D. (faster or slower) than benzene.arrow_forwardPart 2. A solution of 6.00g of substance B in 100.0mL of aqueous solution is in equilibrium, at room temperature, wl a solution of B in diethyl ether (ethoxyethane) containing 25.0 g of B in 50.0 mL 9) what is the distribution coefficient of substance B b) what is the mass of B extracted by shaking 200 ml of an aqueous solution containing 10g of B with call at room temp): i) 100 mL of diethyl ether ii) 50ml of diethyl ether twice iii) 25ml of diethyl ether four timesarrow_forward
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