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× 1. Catalog 2. Science 3. Chemistry practice problems of chap4_5.doc Only available on studygur Topic: Download(s): 0 Pages: 3 Published: 26 октября 2016 READ FULL DOCUMENT Please sign up to read full document. TEXT PREVIEW CHEM 1411 PRACTICE PROBLEMS II (Chapters 4, 5) Multiple Choices: Select one best answer. Q1-11: Chapter 4; Q12-22: Chapter 5; 1. Which of the following is a weak electrolyte? (a) barium hydroxide solution (b) He gas (c) ammonia solution (d) potassium chloride (e) hydrochloric acid Hint: p.p. 120-121. Weak electrolytes include molecular compounds that are weak acids and weak bases. Strong electrolytes include ionic compounds and molecular compounds that are s trong acids. Nonelectrolytes include all molecular compounds except strong acids, weak acids and w eak bases. 2. Which of the following is insoluble in water? (a) (NH4)2CO3 (b) NaI (c) K2SO4 Hint: Refer to p.p.123, Table 4.2 (d) PbI2 (e) Ba(OH)2 3. Which of the following is wrong concerning a net-ionic equation? (a) S2-(aq) + Cu2+(aq) CuS(s) (b) 2NaOH(aq) + MgSO4(aq) Mg(OH)2(s) + Na2SO4(aq) (c) SO42-(aq) + Pb2+(aq) PbSO4(s) (d) CO32-(aq) + Ca2+(aq) CaO + CO2(g) 2+ 2+ (e) Fe(s) +Cu (aq) Fe + Cu(s) Hint: p.p. 124-126. Examples 4.1 & 4.2. Check with the solubility table for the produ cts, which must be solids (s), liquids (l) or gases (g). 4. Which of the following is a Brǿnsted acid? (a) HCO3(b) CH3COO(c) S2(d) SO4(e) NO3Hint: p. 128. Example 4.3. Definition. Brǿnsted acids must contain available (releasable) hydrogen ions while the Brǿnsted bases must be able to accept hydrogen ions. 5. Which of the following underlined atoms contains the oxidation number as -1? (a) Cs2O (b) CaC2 (c) SO42(d) PtCl42(e) Na2O Hint: p.p. 134-137. Rules. Example 4.5. 6. Which of the following redox reactions will occur according to activity series? (a) Cu(s) + 2HCl(aq) CuCl2(aq) + H2(g) (b) Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s) (c) Fe(s) + ZnSO4(aq) FeSO4(aq) + Zn(s) (d) 2Al(s) + 3Ca(NO3)2(aq) 2Al((NO3)3(aq) + 3Ca(s) Hint: Activity table, Fig. 4.6, p. 139. Any metal will be oxidized by the metal ion b elow it in the activity series. 7. Which of the following is not the redox reaction? (a) 2HCl Cl2 + H2 (b) H2 + O2 H2O2 (c) CaCO3 CaO + CO2 (d) 2HNO3 + Mg Mg(NO3) 2+ H2 (e) Cl2+NaBrNaCl+Br2 Hint: Redox reaction involves electron transfer, which leads to change of oxidation n umber of the elements. You need first to figure out the oxidation number of each species. The reac tion that does not have change in oxidation number of elements is not a redox reaction. 8. How many grams of KOH are present in 35.0 mL of a 5.50 M solution? (a) 5.5 (b) 10.8 (c) 15.7 (d) 17.8 (e) 21.3 Hint: p.p. 142-146. Example 4.6. In order to get grams, we need to get moles first. M =n/V, n=MV=5.50M*(35.0/1000)L=0.193mole, m=nM=0.193mole*56.1g/mole=10.8g 9. What is the molarity (M) of a solution made by 5.85 g NaCl in 200.0 mL of solution ? (a) 0.4 (b) 0.5 (c) 0.8 (d) 0.9 (e) 1.4 Hint:molarity=n/V, n=5.85g/(58.44g/mole)=0.1mole, molarity= 0.1mole/(200/1000)L=0.5M 1 10. What is the final concentration (in M) of a solution when water is added to 25.0 mL of a 0.866 M KNO3 solution until the volume of the solution is exactly 500.0 mL? (a) 0.0252 (b) 0.0368 (c) 0.0117 (d) 0.0534 (e) 0.0433 Hint: p.p. 146-148. Similar to the reverse question of Example 4.8. MiVi=MfVf, Mf=MiVi/Vf= 25.0ml*0.866M/500.0ml=0.0433M 11. What volume (in mL) of a 0.500 M HCl solution is needed to neutralize 10.0 mL of a 0.2000 M Ba(OH)2 solution? (a) 8.00 (b) 4.00 (c) 2.00 (d) 1.00 (e) 0.50 Hint: p.p. 150-153. Solution stoichioimetry. Example 4.11. Short-cut formula of acidbase neutralization: ia x Ma x Va = ib x Mb x Vb where a refers to acid and b refers to ba se. ia refers to number of H in the chemical formula of acid and ib refers to the number of OH in the chemica l formula of base. 12. What volume (in L) does a sample of air occupy at 6.6 atm when 1.2 atm, 3.8 L of air is compressed? (a) 0.34 (b) 0.57 (c) 0.69 (d) 0.77 (e) 0.86 Hint: Boyle’s law. p.p. 175-178. Equation (5.2). PV= constant. P1V1=P2V2, V2= P1V1/P2=1.2atm*3.8L/6.6atm = 0.69L 13. What is the final temperature (in K), under constant-pressure condition, when a s ample of hydrogen gas initially at 88oC and 9.6 L is cooled until its finial volume is 3.4 L? (a) 31 (b) 68 (c) 94 (d) 128 (e) 260 Hint: Charles’s Law. p.p. 178-180. Equation (5.4). V/T= constant, V1/T1=V2/T2, T2=V2T1/V1= 3.4L*(88+273.15)K/9.6L=128K 14. What is the volume (in L) of 88.0 g of carbon dioxide gas at STP? (a) 44.8 (b) 53.7 (c) 62.1 (d) 74.6 (e) 83.2 Hint: Ideal gas law. p.p. 181-182. Equation (5.8). Example 5.3. T=0◦C, P=1atm, n=m/M=88.0g/(44g/mole)= 2.0 mole, PV=nRT, V=nRT/P= 2.0mole*0.082 L• atm / (mol • K)*273.15K/1atm=44.8L 15. What is the pressure of the gas (in atm) when 5.0 moles of CO gas are present in a container of 20.0 L at 27 oC? (a) 3.25 (b) 6.15 (c) 7.40 (d) 9.30 (e) 10.55 Hint: PV=nRT, P=nRT/V= 5.0mole*0.082 L• atm / (mol • K)*(27+273.15)K/20.0L=6.15atm 16. A gas evolved during the fermentation of glucose (wine making) has a volume of 0. 78 L at 20.1 oC and 1.00 atm. What was the volume (L) of this gas at the fermentation temperature of 36.5 oC and 2.00 atm pressure? (a) 0.41 (b) 0.82 (c) 1.43 (d) 2.67 (e) 3.54 Hint: Combined gas law. p. 184 equation (5.10). PV/T=Constant, P1V1/T1=P2V2/T2; V2=P1V1T2/T1P2=1.00atm*0.78L*(36.5+273.15)K/[(20.1+273.15)K*2.00atm]=0.41L 17. What is the molar mass (g/mol) of 7.10 grams of gas whose volume is 5.40 L at 741 torr and 40 oC? (a) 34.6 (b) 70.3 (c) 86.2 (d) 94.6 (e) 102.3 Hint: p. 187. Equation (5.12). Example 5.9. M=dRT/P=mRT/VP=7.10g*0.082 L• atm / (mol • K) *(40+273.15)K /[(741/760)atm*5.40L]= 34.6 18. What is the density of HCl gas in grams per liter at 700 mmHg and 25 oC? (a) 0.54 (b) 1.37 (c) 2.24 (d) 2.97 (e) 3.57 Hint:p.p.186, d=m/V=PM/RT=(700/760)atm*36.45g/mole/[0.082 L• atm / (mol • K)*(25+273. 15)K] =1.37g/L 19. A compound has the empirical formula as SF4. At 20 oC, 0.100 gram of the gaseous compound occupies a volume of 22.0 mL and exerts a pressure of 1.02 atm. What is the molecular formula of the gas? 2 (a) SF4 (b) SF6 (c) S2F10 (d) S4F16 (e) S5F20 Hint: p. 211 (5.50) or p. 187. Equation (5.12). Example 5.9. In order to get the mole cular formula, we need to get the molar mass of the compound first. M=dRT/P=mRT/VP =0.100g*0.082 L• atm / (mol • K)*(20+273.15)K/[(22.0/1000)L*1.02atm]=107.1g/mole. The molar mass of the empirical formula SF4 is 32.07+4*19.00=108.07g/mole. This is very similar to the calculated value 107.1g/mole. So the molecular formula is SF4. 20. The combustion process for methane, major component of natural gas, is CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) If 15.0 moles of methane are reacted, what is the volume of carbon dioxide (in L) pro duced at 23.0 oC and 0.985 atm? (a) 369.8 (b) 430.7 (c) 510.8 (d) 630 (e) 720 Hint: Gas stoichiometry: p.p. 190-192. Example 5.11. This question is taken from p. 2 11 (5.52). x 1CO2/1CH4 15 mole CH4 ---------------- 15 mole CO2 PV=nRT, V=nRT/P=15mole*0.082 L• atm / (mol • K)*(23.0+273.15)K/0.985atm=369.8L 20. In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: C6H12O6(s) 2 C2H5OH(l)) + 2 CO2(g) If 5.97 g of glucose are reacted and 1.44 L of carbon dioxide gas are collected at 29 3 K and 0.984 atm, what is the percent yield of the reaction? (a) 89.4 % (b) 76.3% (c) 65.9% (d) 56.2% (e) 47.6% Hint: Gas stoichiometry: p.p. 190-192. This question is taken from p. 212 (5.54).%Y=R eal yield/theoretical yield; theoretical yield can be obtained from reaction of glucose /180g/mole *2CO2/1 C6H12O6 C6H12O6 5.97g ------------------0.033mole----------------------0.066mole---------- Real yield can be obtained though the collection of CO2,PV=nRT, n = PV/RT= 0.984atm*1.44L/[0.082 L• atm / (mol • K)*293K]=0.059mole; %Y= 0.059/0.066=0.894 =89.4% 21. A mixture of gases contains 0.31 mol CH4, 0.25mol C2H6 and 0.29 mol C3H8. The tot al pressure is 1.50atm. Calculate the partial pressure of the CH4, C2H6 and C3H8 gases (a) 0.23,0.35,0.56atm (b) 0.54,0.44,0.51atm (c) 0.54,0.44,0.51atm (d) 0.54,0.44,0.51atm (e) 0.54,0.44,0.51atm Hint : group practice problem Chap5, Q5.63 end of chapter problems. P i=PTXi, nT= 0.31+0.25+0.29=0.85mole; Xi=ni/nT; P CH4= 1.50atm*(0.31mole/0.85mole)=0.54atm; P C2H6 = 1.50atm*(0.2 mole/0.85mole)=0.44 atm; P C3H8= 1.50atm*(0.29mole/0.85mole)=0.51atm; 3 SIGN UP - IT'S FREE × Get access to the full document Please signup to get access to the original document READ FULL DOCUMENT Just to see transcript Similar documents practice problems of chap5 in class summer 2010.doc ... 13. What is the final temperature (in K), under constant-pressure condition, when a sample of hydrogen gas initially at 88oC and 9.6 L is cooled until its finial volume is 3.4 L? (a) 31 (b) 68 (c) 94 (d) 128 (e) 260 Hint: Charles’s Law. p.p. 178-180. Equation (5.4). V/T= constant, V1/T1=V2/T2, T2=V2 ... doc | 48,1 kB | 1 pages QuizBowl ... hydrogen gas initially at 88oC and 9.6 L is cooled until its finial volume is 3.4 L? (a) 31 (b) 68 (c) 94 (d) 128 (e) 260 Hint: Charles’s Law. p.p. 178-180. Equation (5.4). V/T= constant, V1/T1=V2/T2, T2=V2T1/V1= 3.4L*(88+273.15)K/9.6L=128K 14. What is the volume (in L) of 88.0 g of carbon dioxide g ... ppt | 200,7 kB | 0 pages THE ANTIOXIDATIVE AND CYTOTOXIC EFFECTS OF HIBISCUS ... Hint: Activity table, Fig. 4.6, p. 139. Any metal will be oxidized by the metal ion below it in the activity series. 7. Which of the following is not the redox reaction? (a) 2HCl Cl2 + H2 (b) H2 + O2 H2O2 (c) CaCO3 CaO + CO2 (d) 2HNO3 + Mg Mg(NO3) 2+ H2 (e) Cl2+NaBrNaCl+Br2 Hint: Redox reac ... pdf | 2,5 MB | 127 pages Chapter Four: Stoichiometry ... (c) CaCO3 CaO + CO2 (d) 2HNO3 + Mg Mg(NO3) 2+ H2 (e) Cl2+NaBrNaCl+Br2 Hint: Redox reaction involves electron transfer, which leads to change of oxidation number of the elements. You need first to figure out the oxidation number of each species. The reaction that does not have change in oxidatio ... pptx | 3,9 MB | 59 pages doc ... need to get the molar mass of the compound first. M=dRT/P=mRT/VP =0.100g*0.082 L• atm / (mol • K)*(20+273.15)K/[(22.0/1000)L*1.02atm]=107.1g/mole. The molar mass of the empirical formula SF4 is 32.07+4*19.00=108.07g/mole. This is very similar to the calculated value 107.1g/mole. So the molecular for ... doc | 338,4 kB | 15 pages KEY ... Hint: p. 211 (5.50) or p. 187. Equation (5.12). Example 5.9. In order to get the molecular formula, we need to get the molar mass of the compound first. M=dRT/P=mRT/VP =0.100g*0.082 L• atm / (mol • K)*(20+273.15)K/[(22.0/1000)L*1.02atm]=107.1g/mole. The molar mass of the empirical formula SF4 is 32. ... docx | 61,7 kB | 5 pages Thousands of people utilize studygur in their educational activities on a daily basis. Our mission is to provide scholars with the maximum diversity of ideas and concepts for essays, term theses, etc. Contacts New York, NY 10001, 4281 Ward Road support @ studygur.com News and updates subscription Our team constantly updates the portal with various new services Keep up with our news! E-mail Submit © 2017 studygur