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Name:_______________________ CHM 111 Exam 1 **************SHOW ALL WORK FOR FULL CREDIT********************* 200 pts 1. ( 24 pts ) Answer the following questions. Be sure to label your units and express your answers with the correct number of significant figures. a. An object weighs 25.5 lbs. How much does it weigh in kg? (1 kg = 2.2046 lbs) 25.5 x 2.2= 56.1kg b. My daughter rides her bike at an average speed of 15.0 km/hr. Convert this speed to ft/second. (1 mile = 1609.3 m, 1 mile=5280 ft) The average speed of 15.0 km/hr is equivalent to 4.1667 m/s, and multiplying by 3.2808 gives us 13.71 ft/s (rounded to two decimal places). c. A solid has a density of 13.8 g/mL. What is the volume of 27.0 grams of this solid? The volume of the solid is 1.96 mL. 2. ( 8 pts ) How many significant figures are in the following measurements a. 0.000340510 g _6 significant figures______ b. 1.045 mL _4 significant figures______ c. 5000 kg _1 significant figures______ d. 653.0 cm _4 significant figures______ 3. ( 12 points ) For each temperature given in the following table, calculate the equivalent temperatures in the other scales. o F o C K -24 -31.1 242.03 9 80.6 27 399.15 197.3 91.85 365 1

Name:_______________________ CHM 111 Exam 1 3 6.022x10 23 molecules/mol 4. ( 24 pts ) Perform the following calculations for C 7 H 12 N 2 O 3 (a) Calculate the molar mass of C 7 H 12 N 2 O 3 . The molar mass of C7H12N2O3 is 172.19 g/mol. (b) How many moles of C 7 H 12 N 2 O 3 are in a 10.0g sample? There are 0.058 moles of C7H12N2O3 in a 10.0 g sample. (c) How many molecules of C 7 H 12 N 2 O 3 are in a 10.0g sample? There are 3.50 x 10^22 molecules of C7H12N2O3 in a 10.0 g sample. (d) How many Carbon atoms are in a 10.0g sample of C 7 H 12 N 2 O 3 ? There are 20 moles of carbon atoms in a 10.0 g sample of C7H12N2O3, which is equivalent to 1.20 x 10^24 carbon atoms. 2

Name:_______________________ CHM 111 Exam 1 5. (18 pts) Fill in the gaps in the following table: Symbol Mg 80 Se 2- Ca Sn Sr 37 Cl - Au Protons 12 34 20 50 36 17 79 Neutrons 12 46 21 68 50 20 118 Electrons 12 36 18 46 36 18 78 Net charge 0 -2 +2 +2 0 -1 +1 6. ( 20 pts ) Give the systematic name of each of the following compounds. Indicate if the compound is ionic or molecular (circle). (a) Sr 3 (PO 4 ) 2 ____________________ ionic molecular (b) HNO 3 _________________acid ionic molecular (c) O 3 F ____________________ ionic molecular (d) Fe(OH) 3 ____________________ ionic molecular (e) SF 6 ____________________ ionic molecular 7. ( 12 pts ) Determine the empirical formula of a compound composed of 32.4% Na, 22.5% S, and 45.1%O by mass. The empirical formula of the compound is Na2SO4. 3

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Name:_______________________ CHM 111 Exam 1 BONUS: Name the compound:____________________ 8. (25 pts) sodium hydroxide reacts with hydrochloric acid to form sodium chloride and water. (a) (5 pts ) Write the balanced chemical equation for this reaction The balanced chemical equation for the reaction between sodium hydroxide and hydrochloric acid is: NaOH + HCl → NaCl + H2O (b) ( 12 pts ) How many grams of sodium chloride form when 25.0 g of hydrochloric acid and 25.0 g of sodium hydroxide are mixed? (Put a box around your final answer) The balanced equation for the reaction between hydrochloric acid and sodium hydroxide is: HCl + NaOH → NaCl + H2O To determine the amount of sodium chloride formed, we need to first determine the limiting reactant. The balanced equation shows that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of NaCl. The number of moles of HCl in 25.0 g can be calculated as: n(HCl) = m(HCl) / M(HCl) = 25.0 g / 36.46 g/mol = 0.686 mol 4

Name:_______________________ CHM 111 Exam 1 The number of moles of NaOH in 25.0 g can be calculated as: n(NaOH) = m(NaOH) / M(NaOH) = 25.0 g / 40.00 g/mol = 0.625 mol Since NaOH is the limiting reactant (it produces fewer moles of product), we can use its mole ratio with NaCl to determine the number of moles of NaCl produced: n(NaCl) = n(NaOH) = 0.625 mol The mass of NaCl produced can be calculated as: m(NaCl) = n(NaCl) x M(NaCl) = 0.625 mol x 58.44 g/mol = 36.5 g Therefore, 36.5 g of sodium chloride form when 25.0 g of hydrochloric acid and 25.0 g of sodium hydroxide are mixed. (c) ( 3pt ) Which compound is the limiting reagent? To determine the limiting reagent, we need to compare the number of moles of each reactant present and their stoichiometric coefficients in the balanced equation. 5

Name:_______________________ CHM 111 Exam 1 From the previous calculation, we found that the number of moles of HCl is 0.686 mol and the number of moles of NaOH is 0.625 mol. The balanced equation shows that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of NaCl. Therefore, the reactant that produces fewer moles of product is the limiting reagent. In this case, NaOH produces 0.625 moles of NaCl, while HCl would produce 0.686 moles of NaCl if it were completely consumed. Since NaOH produces fewer moles of NaCl, it is the limiting reagent. (d) ( 5 pts ) After performing the experiment, 15.0 g of sodium chloride are recovered. Calculate the percent yield. The theoretical yield of sodium chloride can be calculated from the balanced equation and the limiting reagent: 1 mol NaOH produces 1 mol NaCl 0.625 mol NaOH produces 0.625 mol NaCl 0.625 mol NaCl x 58.44 g/mol = 36.5 g NaCl (theoretical yield) The percent yield can be calculated as: percent yield = (actual yield / theoretical yield) x 100% Substituting the given values, we get: percent yield = (15.0 g / 36.5 g) x 100% = 41.1% Therefore, the percent yield of sodium chloride is 41.1%. 9. ( 15 pts ) Given the following systematic names, write the formula for each compound. 6

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Name:_______________________ CHM 111 Exam 1 (a) dinitrogen tetroxide __ N2O4_________________ (b) cobalt(II) bromide __ CoBr2_________________ (c) methane ____ CH4_______________ (d) ammonium sulfate ____(NH4)2SO4_______________ (e) carbonic Acid __ H2CO3_________________ 10. ( 10 pts ) When 10.00 g of Fe(NO 3 ) 3  xH 2 O hydrate is heated to 250 o C, all the water of hydration is lost, leaving 5.99 g of Fe(NO 3 ) 3 . What is the value of x? To determine the value of x in Fe(NO3)3•xH2O, we need to calculate the number of moles of Fe(NO3)3 and H2O in the original sample and use the mole ratio to find x. The molar mass of Fe(NO3)3 is 241.86 g/mol, and the molar mass of H2O is 18.02 g/mol. First, we need to calculate the number of moles of Fe(NO3)3 in the sample: 10.00 g Fe(NO3)3•xH2O x (1 mol Fe(NO3)3 / 241.86 g) = 0.0413 mol Fe(NO3)3 Next, we need to calculate the number of moles of Fe(NO3)3 in the anhydrous sample: 5.99 g Fe(NO3)3 x (1 mol Fe(NO3)3 / 241.86 g) = 0.0248 mol Fe(NO3)3 The difference in moles between the original sample and the anhydrous sample is due to the loss of water: 7

Name:_______________________ CHM 111 Exam 1 0.0413 mol Fe(NO3)3 - 0.0248 mol Fe(NO3)3 = 0.0165 mol H2O Finally, we can use the mole ratio between Fe(NO3)3 and H2O to find x: 0.0165 mol H2O x (1 mol Fe(NO3)3 / 1 mol H2O) = 0.0165 mol Fe(NO3)3 This means that there is 0.0165 mol Fe(NO3)3 in 10.00 g Fe(NO3)3•xH2O. The molar mass of Fe(NO3)3•xH2O is: 10.00 g Fe(NO3)3•xH2O / 0.0165 mol Fe(NO3)3•xH2O = 605.05 g/mol The molar mass of Fe(NO3)3 is: 241.86 g/mol The molar mass of H2O is: 18.02 g/mol Therefore, the formula for the hydrate is Fe(NO3)3•16H2O. 8

Name:_______________________ CHM 111 Exam 1 11. ( 8 pts ) The empirical formula of a compound is C 5 H 4 N 4 O 7 . The compound has a molar mass of 464.26 g/mol; determine the molecular formula. To determine the molecular formula of the compound, we need to know the empirical formula mass, which is the sum of the atomic masses of the empirical formula: Empirical formula mass = (5 x 12.01 g/mol) + (4 x 1.01 g/mol) + (4 x 14.01 g/mol) + (7 x 16.00 g/mol) = 196.07 g/mol Next, we can calculate the ratio of the molecular formula mass to the empirical formula mass: Ratio = Molecular formula mass / Empirical formula mass = 464.26 g/mol / 196.07 g/mol = 2.37 The ratio tells us how many empirical formula units are present in the molecular formula. We can round this to the nearest whole number to get the subscript for each element in the molecular formula: C5H4N4O7 x 2 = C10H8N8O14 Therefore, the molecular formula of the compound is C10H8N8O14. 12. ( 8 pts ) Identify each of the following as an intensive or extensive property a. Volume intensive extensive b. freezing point intensive extensive c. Mass intensive extensive d. Density intensive extensive Multiple Choice 16 points 1. 45 m/s = __________ km/hr a. 1.6 × 10 5 b. 2.7 c. 0.045 9

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Name:_______________________ CHM 111 Exam 1 d. 1.6 × 10 2 e. 2.7 × 10 3 2. A combination of sand, salt, and water is an example of a __________. a. homogeneous mixture b. compound c. pure substance d. heterogeneous mixture e. solid 3. What is the correct name for the following compound: Fe 2 (SO 4 ) 3 a. iron(II) sulfate b. iron(III) sulfate c. iron(III) sulfite d. iron(II) sulfide e. iron(III) sulfide 4. How many grams of lead (II) oxide will be produced by the decomposition of 2.50 g of lead (II) carbonate? PbCO 3 (s) → PbO (s) + CO 2 (g) a. 2.61 b. 0.41 c. 2.50 d. 0.00936 e. 2.09 10