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Chemistry 30 Chem. 20 Review Answers 1. In the following chart you will find partial information about several atoms and ions. Fill in the missing information using your table of atomic weights. (You are to determine the atomic mass from the information in the question table, not from your atomic masses table). Name Symbol Atomic Atomic # of # of # of Number Mass Protons Neutrons Electrons calcium Ca 20 40 20 20 20 atom chromic Cr3+ 24 52 24 28 21 ion chlorine Cl 17 35 17 18 17 atom oxide ion O2- 8 16 8 8 10 calcium ion Ca2+ 20 40 20 20 18 nitride N3- 7 40 7 7 10 ion 2. Write the correct chemical (empirical) formulas for the following compounds using the information in your ion sheets. a) beryllium chloride f) ferrous hydroxide BeCl2 Fe(OH)2 b) gallium iodide g) diphosphorous pentoxide GaI3 P2O5 c) calcium hydroxide h) lead (II) bicarbonate Ca(OH)2 Pb(HCO3)2 d) phosphoric acid i) ammonium phosphate H3PO4 (NH4)3PO4 e) ferric hydroxide Fe(OH)3 3. Calculate the numbers of moles of atoms present in the following samples: a) 3.6g C b) 54g Al # of moles = sample mass molar mass a) moles C = 3.6g 12g/mol = 0.3 mol b) moles Al = 54g 27g/mole = 2.0 mol 4. Calculate the molecular weights (molar masses) of the following compounds: a) HCl c) ammonium dichromate b) SrSO4 d) aluminum carbonate a) HCl 1 - H is 1 x 1.008 = 1.008 1 - Cl is 1 x 35.5 = 35.5 36.508 7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012 Page 1 of 11 Chem. 20 Review Answers - pg. 2 b) SrSO4 1 - Sr is 1 x 87.6 = 87.6 1 - S is 1 x 32.1 = 32.1 4 - O is 4 x 16.00 = 64.00 183.7 c) (NH4)2Cr2O7 2 - N is 2 x 14.01 = 28.02 8 - H is 8 x 1.008 = 8.064 2 - Cr is 2 x 52.0 = 104.0 7 - O is 7 x 16.00 = 112.00 252.084 d) Al2(CO3) 3 2 - Al is 2 x 27.0 = 54.0 3 - C is 12.01 = 36.03 9 - O is 9 x 16.00 = 144.00 234.03 5. Calculate the numbers of moles of molecules present in the following samples: a) 4.4g CO2 b) 34g AgNO3 The general strategy is to divide sample mass by molar mass to determine numbers of moles. Before this can be done you must calculate the molar mass (molecular weight) of each compound as in #4. a) # of moles = 4.4g 44g/mol = 0.1mol b) # of moles = 34g 170g/mol = 0.2mol 6. Calculate the mass in grams of one (1) water molecule. The chemical formula water, for those of you who forgot, is H2O. 1 mole of anything contains Avogadro’s Number of particles which is 6.02 x 1023. The mass of 1 mole of water is 18.016g (using same procedure as in #4 & #5). From this we can write: 6.02 x 1023 water molecules have a mass of 18.016g. 1 water molecule has a mass of: 18.016g 6.02 x 1023 = 2.99 x 1023g 7. Calculate the number of oxygen atoms contained in 15g of calcium carbonate. The formula for calcium carbonate is CaCO3, which tells us that 1 molecule of the compound contains 3 oxygen atoms. , # of CaCO3 molecules 3 should give us the number of oxygen atoms. The number of CaCO3 molecules should be the number of moles Avogadro’s Number. Moles of CaCO3 should be sample mass molar mass. # of moles CaCO3 = 15g 100g/mol = 0.15mol The remaining two steps can be shown in one calculation: # of O atoms = 0.15mol 6.02 x 1023 3 = 2.71 x 1023 O atoms 8. Balance the following equations: a) 1 C7H16 + 11 O2 7 CO2 + 8 H2O b) 4 Fe + 3 O2 2 Fe2O3 7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012 Page 2 of 11 Chem. 20 Review Answers - pg. 3 c) 3 NO2 + 1 H2O 2 HNO3 + 1 NO d) 4 NaHCO3 + 2 HCl 1 Na2CO3 + 2 NaCl + 3 CO2 + 3 H2O 9. Write a balanced equation for the complete combustion (burning) of propane (C3H8). Combustion means combining with oxygen gas (O2). The products of combustion are carbon dioxide and water. For the complete combustion of 66g of propane, calculate: a) the mass of oxygen required b) the mass of CO2 produced c) the mass of H2O produced. C3H8 + 5 O2 3 CO2 + 4 H2O Assuming the coefficients represent moles of each substance, we can write the masses involved by determining molecular weights and multiplying them by the coefficients. C3H8 + 5 O2 3 CO2 + 4 H2O 44g 160g 132g 72g (5 x 32) (3 x 44) (4 x 18) a) the mass of oxygen required This first method is assuming you can do grocery store arithmetic. 44g C3H8 requires 160g O2 1 g C3H8 requires 160g 44g O2 66g C3H8 requires (160 44) 66 = 240g O2 This could also be solved using ratios. Set up a ratio of mass C3H8 : mass O2 for the balanced equation. Make this equal to the ratio of the same two substances in our new situation. Balanced Equation: New Situation: mass C3H8 = mass C3H8 mass O2 mass O2 44g = 66g 160g mass O2 When two ratios or fractions are equal to each other, the cross products are equal. Working these out we get: mass O2 44g = 160g 66g mass O2 = 160g 66g 44g mass O2 = 240g b) the mass of CO2 produced 44g C3H8 rproduces 132g CO2 1 g C3H8 produces 132g 44g CO2 66g C3H8 produces (132 44) 66 = 198g CO2 OR: Balanced Equation: New Situation: 7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012 Page 3 of 11 Chem. 20 Review Answers - pg. 4 mass C3H8 = mass C3H8 mass CO2 mass CO2 44g = 66g 132g mass CO2 mass CO2 44g = 132g 66g mass CO2 = 132g 66g 44g mass CO2 = 198g c) the mass of H2O produced. 44g C3H8 rproduces 72g H2O 1 g C3H8 produces 72g 44g H2O 66g C3H8 produces (72 44) 66 = 108g H2O OR: Balanced Equation: New Situation: mass C3H8 = mass C3H8 mass H2O mass H2O 44g = 66g 72g mass H2O mass O2 44g = 72g 66g mass H2O = 72g 66g 44g mass H2O = 108g 10. Calculate the molar volume of any gas at 200C and 850mmHg pressure. The molar volume of any gas at S.T.P. is 22.4L. S.T.P. is standard temperature and pressure, which is 0C (273K)and 1 atmosphere pressure (760mmHg). Our new conditions are 200C (473K) and 850mmHg. The gas formula which deals with pressure, volume and temperature together is the combined gas law. Substituting P1V1 = P2V2 our known values into this T1 T2 relationship we get: P1V1 = P2V2 T1 T2 760mmHg 22.4L 850mmHg V2 273K 473K Once again, our cross products should be equal. 760mmHg 22.4L = 850mmHg V2 273K 473K 760mmHg 22.4L = V2 473K 850mmHg 273K 7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012 Page 4 of 11 Chem. 20 Review Answers - pg. 5 34.7L = V2 11. A gas from an unmarked cylinder was found to have a density of 1.20g/L.at 25 C and one standard atmosphere (760mmHg) pressure. What must be the molecular weight of the gas? The density tells us the mass of 1L is 1.20g. The molecular weight would be the mass of however many litres there are in one mole at the stated conditions. We know molar volume is 22.4L at S.T.P., so the first thing we do is determine the molar volume at 25C and 760mmHg. Since only the temperature is different from S.T.P., we use the formula: V1 = V2 T1 T2 Substituing the new temperature into the equation and solving for the new volume we get: V1 = V2 T1 T2 22.4L = V2 273K 298K V2 273K = 22.4L 298K V2 = 22.4L 298K 273K V2 = 24.45L Therefore, since 1 mole of gas at these conditions is 24.45L and 1L of the gas has a mass of 1.20g, the molecular weight is: 24.45 L 1.20 g = 29.34 g mol L mol 12. 64g of oxygen gas and 64g of sulfur dioxide gas are placed into a closed container and found to exert a total pressure of 1,000mmHg. Calculate the partial pressure of each gas. The law of partial pressures states that each gas in a mixture exerts a partial pressure in proportion to the number of moles of gas present. We must first determine the moles of each gas present. Moles oxygen (O2) = 64g 32g g/mol = 2 moles moles sulfur dioxide(SO2) = 64g 64g/mol = 1 mole there are 3 moles of gas in the container. Since oxygen accounts for 2/3 of the moles, then it also accounts for 2/3 of the pressure. Similarly, the sulfur dioxide accounts for 1/3 of the moles and 1/3 of the total pressure. Pressure O2 = 2/3 1,000mmHg = 666.7mmHg Pressure S O2 = 1/3 1,000mmHg = 333.3mmHg 13. A steel cylinder is evacuated with a pump until the pressure gauge reads zero. Then, 48g of O2(g) is admitted into the container resulting in a pressure reading of 15 p.s.i. Next, 90g of gas X is added to the 7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012 Page 5 of 11 Chem. 20 Review Answers - pg. 6 container and the pressure gauge reads 30 p.s.i. What must be the molecular weight of gas X? Since the pressure of the oxygen alone is 15psi and the pressure of the two together is 30psi, the pressure of the unknown alone would be 15psi. Since these measurements were taken in the same steel cylinder (equal volumes) at the same temperature, and each of them alone would exert the same pressure, this situation fits Avogadro’s Hypothesis nicely: “Equal volumes of different gases measured at the same temperature and pressure contain equal numbers of molecules”. 48g of oxygen contains the same number of moles as 90g of the unknown. Moles oxygen present is: 48g 32g/mol = 1.5mol. If 1.5mol of the unknown has a mass of 90g, then 1 mol mas a mass of: 90g 1.5mol = 60g/mol. We could also reason as follows: since the two gases have equal numbers of moles, we can say sample mass molar mass for the two will be equal. We can then write: Oxygen Unknown: Sample mass = Sample mass Molar mass Molar mass 48g = 90g 32g/mol molar mass molar mass 48 = 90 32 molar mass = (90 32) 48 = 60g/mol 14. A cylinder of propane is filled so that ¾ of the cylinder is liquid propane and the top ¼ is gaseous propane. The pressure in the tank is measured to be 500kPa. After some time enough propane is used so that the cylinder is ½ liquid and ½ gas. Assuming that the temperature has not changed, what would be the reading on the pressure gauge? Explain your answer. The pressure in the tank will still be 500kPa. We are measuring the vapor pressure, and as long as there is some liquid and some gas, the vaspor pressure will be the same. The only way we could change the pressure would be to change the temperature. 15. If you were to dissolve 40g of calcium bromide into enough water to give 500mL of solution, a) What would be the concentration (molarity) of the calcium bromide solution? Molarity is the moles of solute present in 1 litre of solution. It can be calculated as: moles solute litres solution. The formula is: M = n V, or M = n V 7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012 Page 6 of 11 Chem. 20 Review Answers - pg. 7 Moles CaBr2 = 40g 200g/mol = 0.2mol M = n = 0.2mol = 0.4M V 0.5L b) What would be the concentration of calcium and bromide ions in the solution? Calcium bromide dissolves in water according to the equation: CaBr2(s) Ca2+(aq) = 2 Br(aq) Since 1 CaBr2 produces 1 Ca2+ and 2 Br , the concentration of the calcium ion is the same as that of the compound, while the concentration of the bromide ion is twice that of the compound. [Ca2+(aq)] =- 0.4M [Br((aq)] = 0.8M 16. How many grams of pure AgNO3 would be contained in 250mL of a 0.5M AgNO3 solution? We need to know the number of moles of silver nitrate first. Rearranging the basic molarity formula we should get: n = M V, or moles equals molarity times volume. Substituting our given values we get: n = M V = 0.5mol/L 0.25L = 0.125mol mass AgNO3 = 0.125mol 169.91g/mol = 21.24g 17. If you had only 10.3g of pure solid NaBr available and you wished to prepare a 0.25M solution of it, what would be the maximum volume of this solution that could be made? Rearranging the basic molarity formula we should get: V = n or Volume = moles M Molarity Before we can use this we need to know the moles of NaBr. Moles NaBr = 10.3g 102.9g/mol = 0.1mol Substituting this into our revised formula we get: V = n = 0.1mol = 0.4L, or 400mL M 0.25mol/L 18. If you took 250mL of a 0.6M HCl solution and added 750mL of water to it, what would be the concentration of the final solution? Since adding water (or taking it out) does not alter the moles of HCl, we can say that moles HCl before the dilution is equal to moles of HCl after dilution. Since moles = molarity x volume, we can derive the formula: M1V1 = M2V2 Substituting our values into this equality we get: 0.6M 0.25L = M2 1.0L M2 = (0.6M 0.25L) 1.0L = 0.15M 19. A student made the following three pairs of mixtures of aqueous ionic solutions and observed which combinations produced precipitates: i) Ba(NO3)2 and Na2SO4 gave a white precipitate ii) Ca(NO3)2 and NaBr gave no precipitate iii) CaCl2 and Na2CO3 gave a white precipitate. 7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012 Page 7 of 11 Chem. 20 Review Answers - pg. 8 a) For each combination that produced a precipitate, indicate what the two possibilities are for the solid product. Ba(NO3)2 has the Ba2+ and NO3 ions, Na2SO4 has Na+ and SO42. The new combinations of + and - ions would be NaNO3 and Ba SO4 CaCl2 has the Ca2+ and Cl ions. The Na2CO3 has the Na+ and the CO32 ions. The two new combinations are NaCl and CaCO3. b) Determine which of the new combinations are precipitates, and which are soluble combinations. Explain your reasoning. Of the two possibilities from i) the NaNO3 can be eliminated since there was no reaction with set ii) which brought the sodium and nitrate ions together. The precipitate has to be Ba SO4. In set iii) the NaCl can be elininated since this is table salt, which as we all should know, is soluble in water. The precipitate has to be CaCO3. c) Write the net ionic equation for the formation of the two precipitates. Ba2+(aq) + SO42(aq) BaSO4(s) Ca2+(aq) + CO32(aq) CaCO3(s) d) Using the information presented so far, indicate for each of the following compounds whether you would expect it to be soluble or insoluble in water. NaNO3 Ba(NO3)2 BaSO4 soluble, no ppt. In soluble, since it is an insoluble, see 19b) reaction ii) which original solution. brings sodium and nitrate ions together. CaBr2 CaCO3 CaSO4 soluble, for same insoluble, see 19b) can’t be determined reason NaNO3 from given information. Se solubility chart, which says it is insoluble. 20. A chemist mixed 18g of carbon, 6g of hydrogen gas and 160g of oxygen gas in a closed container. She then ignited it electrically so that the hydrogen united with oxygen to form water. The carbon then burned to form carbon dioxide. The reaction proceeded until one or more of the reactants was completely used up. a) Write balanced equations to represent the two reactions happening.. 2H2 + O2 2 H2O C + O2 CO2 b) Which of the reactants would still be left over, and what mass of it would there be? The question indicates that the oxygen reacts with the hydrogen first, then any oxygen left over reacts with the carbon. We have to treat this as two separate reactions. 2H2 + O2 2 H2O 7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012 Page 8 of 11 Chem. 20 Review Answers - pg. 9 C + O2 CO2 For the reaction of hydrogen, the equation says that 4g hydrogen needs 32g oxygen. Therefore, 6g hydrogen needs: (32 4) 6 = 48g oxygen. This can happen since we have 160g of oxygen gas. Following this reaction there will be: 160g 48g = 112g of oxygen left. For the reaction of carbon, the equation says 12g carbon needs 32g oxygen. Therefore, 18g carbon needs: (32 12) 18 = 48g oxygen. This can also happen since there are 112g of oxygen left. Following this reaction there will be: 112g 48g = 64g of oxygen left. c) What mass of water would be formed? From the equation, 4g of hydrogen produces 36g water, so 6g hydrogen should make: (36 4) 6 = 54g water. d) What volume of carbon dioxide gas would be formed at 30C and 150kPa pressure? Steps to be taken: 1) determine mass of carbon dioxide produced 2) determine moles of carbon dioxide this is 3) determine molar volume under the conditions specified 4) multiply moles x molar volume. 1) 12g carbon makes 44g carbon dioxide, therefore 18g carbon makes (44 12) 18 = 66g carbon dioxide 2) moles carbon dioxide = 66g 44g/mol = 1.5mol 3) molar volume at S.T.P. is 22.4L. This is 0C (273K) and 101.325kPa. Our new conditions are 30C (303K) and 150kPa. Using the combined gas law we get a new molar volume of: P1V1 = P2V2 T1 T2 101.325kPa 22.4L = 150kPa V2 273K 303K V2 = 101.325kPa 22.4L 303K 150kPa 273K V2 = 16.8L We could do a similar calculation using the value for molar volume a S.A.T.P., which is 24.8L at 25C and 100kPa. P1V1 = P2V2 T1 T2 100kPa 24.8L = 150kPa V2 298K 303K 7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012 Page 9 of 11 Chem. 20 Review Answers - pg. 10 V2 = 100kPa 24.8L 303K 150kPa 298K V2 = 16.8L Since we have 1.5mol, the volume of gas would be: 1.5mol 16.8L/mol = 25.2L e) What would be the total mass of all materials in the container at the end of the reaction? The law of conservation of mass states that total mass of products will equal total mass of reactants. Total mass of reactants was: 18g + 6g + 160g = 184g carbon hydrogen oxygen f) After the material had cooled down to its initial temperature, would the gas pressure in the container have changed? Which way? Explain your answer. To judge any change in presssure we must know total moles of gas before and after the reaction. If we rewrite our equation to show the states of reactants and products at room temperature we get: 2 C(s) + 2H2(g) + 3O2(g) 2 H2O(l) + 2CO2(g) On the reactant side we had 6g 2g/mol = 3mol H2 and 160g 32g/mol = 5mol O2 for a total of 8 moles of gas molecules. Assuming the water cools and condenses to the liquid state, for products we have 1.5mol CO2 (see part d). Before the reaction we had 8 moles of gas, after we have 1.5 moles. The gas pressure in the container will go down. 21. Calcualte the anount of heat necessary to raise the temperature of 450g of water from 10C to 75C. The formula you need is: Q = mcT, where Q = quantity of heat m = mass in grams c = specific heat capacity ( in cal/gC) T = change in temperature The specific heat capacity of water is 1 cal/gC. Q = mcT, where m = 450g c = 1cal/gC T = 75C 10C = 65C Q = 450g 1cal/gC 65C Q = 29,250cal or 29.25kcal 22. Calculate the amount of heat released when 75g of steam at 100C condenses into water at 100C. The heat of vaporization of water is 9.7kcal/mole. The formula you need is: Heat = Amount of substance heat of fusion. The ‘amount of substance’can be expressed as a mass in grams, or as a number of moles. If you wish to work in grams, you must first 7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012 Page 10 of 11 Chem. 20 Review Answers - pg. 11 determine the heat of vaporization in cal/g. If you wish to work in moles, you must first determine the moles of the substance in the sample. Working in moles: Moles water = 75g 18g/mol = 4.16mol Heat = 4.2mol 9.7kcal/mol = 40.352kcal or 40,352cal Working in grams: Heat of vaporization in cal/g: 9,700cal/mol 18g/mol = 539cal/g Heat = 75g 539cal/g = 40,425cal These answers are a bit different, but if we round them off to three significant figures, we get 40.4kcal for both. 7ced7cdf-1e7e-46b1-9803-35d4c4121be4.docAugust 9, 2012 Page 11 of 11