Tin reacts with fluorine to form two different compounds, A and B. Compound A contains 38.5 g of tin for each 12.3 g of fluorine. Compound B contains 56.5 g of ...
Chemistry I Sem II exam On a seperate sheet of paper provide a short answer for three of the following Questions. Please State the Question and answer in full sentences with correct sentence structure, spelling and punctuation. 1. Tin reacts with fluorine to form two different compounds, A and B. Compound A contains 38.5 g of tin for each 12.3 g of fluorine. Compound B contains 56.5 g of tin for each 36.2 g of fluorine. What is the lowest whole-number mass ratio of tin that combines with a given mass of fluorine? 2. How many representative particles are in 1.45 g of a molecular compound with a molar mass of 237 g? 3. Find the mass in grams of 3.10
10
molecules of F .
4. Find the number of moles of argon in 607 g of argon. 5. Find the mass, in grams, of 1.40
10
molecules of N .
6. What is the percent composition of NiO, if a sample of NiO with a mass of 41.9 g contains 33.1 g Ni and 8.8 g O? 7. What is the percent by mass of hydrogen in aspirin, C H O ? 8. Calculate the molecular formulas of the compounds having the following empirical formulas and molar masses: C H , 58 g/mol; CH, 78 g/mol; and HgCl, 236.1 g/mol. 9. Balance the everyday equation described in the following sentence. "A tricycle is made of a handle, a body, three wheels, and two axles. H B W A HBW A 10. Complete and balance the following equation. Cd(NO ) NH Cl 11. Balance the following equation. NaClO NaCl O 12. Balance the following equation. Mg H PO Mg (PO ) H 13. Balance the following equation. (NH ) CO NaOH Na CO
NH
H O
14. Balance the following equation. C H
O
CO
H O
15. Balance the following equation. Complete the equation first, if necessary. Ba H O Ba(OH) H 16. Balance the following equation. Au O Au O 17. Balance the following equation. Na PO ZnSO Na SO
Zn (PO )
18. Complete and balance the following equation. Al Cl
19. Complete and balance the following equation. CH
O
CO
20. Complete and balance the following equation. Fe (SO ) Ba(OH) 21. Balance the following equation. Indicate whether combustion is complete or incomplete. C H O CO H O 22. Balance the following equation. Indicate whether combustion is complete or incomplete. 2C H OH O CO H O 23. Write a balanced net ionic equation for the following reaction. H PO (aq) Ca(OH) (aq) Ca (PO ) (aq) H O(l) 24. Complete and balance the following equation: K PO BaCl 25. What is a pressure of 0.520 atm equal to in mm of Hg? 26. What is a pressure of 622 mm Hg equal to in atm? 27. At what temperature do particles theoretically have no kinetic energy? 28. The vapor pressure of 10 mL of ethanol at 20 C is 5.85 kPa. What is the vapor pressure of 20 mL of ethanol at the same temperature? 29. What is the pressure (in atm) at the normal boiling point of water? 30. What is the angle measurement in cubic, tetragonal, and orthorhombic crystal systems? 31. The volume of a gas is 250 mL at 340.0 kPa pressure. What will the volume be when the pressure is reduced to 50.0 kPa, assuming the temperature remains constant? 32. A balloon filled with helium has a volume of 30.0 L at a pressure of 100 kPa and a temperature of 15.0 C. What will the volume of the balloon be if the temperature is increased to 80.0 C and the pressure remains constant? 33. A gas has a volume of 590 mL at a temperature of –55.0 C. What volume will the gas occupy at 30.0 C? 34. A rigid container of O has a pressure of 340 kPa at a temperature of 713 K. What is the pressure at 273 K? 35. A 10-g mass of krypton occupies 15.0 L at a pressure of 210 kPa. Find the volume of the krypton when the pressure is increased to 790 kPa. 36. A gas has a pressure of 710 kPa at 227 C. What will its pressure be at 27 C, if the volume does not change? 37. A gas occupies a volume of 140 mL at 35.0 C and 97 kPa. What is the volume of the gas at STP? 38. A gas storage tank has a volume of 3.5 10 m when the temperature is 27 C and the pressure is 101 kPa. What is the new volume of the tank if the temperature drops to –10 C and the pressure drops to 95 kPa? 39. How many moles of N are in a flask with a volume of 250 mL at a pressure of 300.0 kPa and a temperature of 300.0 K? 40. The gaseous product of a reaction is collected in a 25.0-L container at 27 C. The pressure in the container is 300.0 kPa and the gas has a mass of 96.0 g. How many moles of the gas are in the container?
41. What is the pressure exerted by 32 g of O in a 22.0-L container at 30.0 C? 42. A mixture of gases at a total pressure of 95 kPa contains N , CO , and O . The partial pressure of the CO is 24 kPa and the partial pressure of the N is 48 kPa. What is the partial pressure of the O ? 43. Use Graham’s law to calculate how much faster fluorine gas, F , will effuse than chlorine gas, Cl , will. The molar mass of F = 38.0; the molar mass of Cl = 70.9. 44. What is the angle between the bonds of a water molecule? 45. At what temperature does liquid water have its maximum density? 46. What is the percentage of water in the hydrate CoCl
6
47. How many grams of copper sulfate pentahydrate (CuSO water?
? 5H O) would you heat to produce 29.8 g of
Chemistry I Sem II exam Answer Section SHORT ANSWER 1. ANS: The mass ratio of tin per gram of fluorine is 2:1. Compound A: 38.5 g Sn/12.3 g F = 3.12 g Sn/1.00 g F Compound B: 56.5 g Sn/36.2 g F = 1.56 g Sn/1.00 g F 3.12 g Sn (Compound A)/1.56 g Sn (Compound B) = 2/1 PTS: 1 DIF: L3 OBJ: 9.5.1 2. ANS: 1.45 g 1.00 mol/237 g 6.02 = 3.68 10 molecules
REF: p. 274 | p. 275
10
PTS: 1 DIF: L2 STA: Ch.Ch.2.3 3. ANS: 3.10 10 molecules 1 mol F /6.02 = 19.6 g F
molecules/1.00 mol
REF: p. 292
10
molecules
OBJ: 10.1.2
38.0 g F /1 mol F
PTS: 1 DIF: L2 REF: p. 297 STA: Ch.Ch.2.3 4. ANS: 607 g Ar 1 mol Ar/39.9 g Ar = 15.2 mol Ar
OBJ: 10.2.1
PTS: 1 DIF: L2 REF: p. 299 OBJ: 10.2.1 STA: Ch.Ch.2.3 5. ANS: 1.40 10 molecules N (1.00 mol N /6.02 10 molecules N ) (28.0 g N /1 mol N ) = 6.51 g PTS: 1 DIF: L3 OBJ: 10.1.2 | 10.2.1 6. ANS: 33.1 g Ni/41.9 g NiO 100% = 79% Ni 8.8 g O/41.9 g NiO 100% = 21% O
REF: p. 291 | p. 297 STA: Ch.Ch.2.3
PTS: 1 DIF: L2 REF: p. 307 STA: Ch.Ch.2.3 7. ANS: 8.00 g H /180 g C H O 100% = 4.44% H
OBJ: 10.3.1
PTS: 1 DIF: L3 REF: p. 307 STA: Ch.Ch.2.3 8. ANS: 58 g/mol/29 g/efm = 2 efm/mol; C H 78 g/mol/13 g/efm = 6 efm/mol; C H 236.1 g/mol/236.1 g/efm = 1 efm/mol; HgCl
OBJ: 10.3.1
PTS: 1 DIF: L3 STA: Ch.Ch.2.3 9. ANS: H + B + 3W + 2A HBW A
REF: p. 312
OBJ: 10.3.3
PTS: 1 10. ANS: CdCl 2NH NO
DIF: L1
REF: p. 325
OBJ: 11.1.3
PTS: 1 11. ANS: 2NaClO
DIF: L2
REF: p. 327
OBJ: 11.1.3
REF: p. 327
OBJ: 11.1.3
REF: p. 327
OBJ: 11.1.3
2NaCl
PTS: 1 12. ANS: 3Mg 2H PO PTS: 1 13. ANS: (NH ) CO
DIF: L2 Mg (PO ) DIF: L2 2NaOH
PTS: 1 14. ANS: C H
PTS: 1 17. ANS: 2Na PO PTS: 1 18. ANS:
Na CO
DIF: L2 3O
PTS: 1 15. ANS: Ba 2H O PTS: 1 16. ANS: 2Au O
3O
3CO
REF: p. 327
OBJ: 11.1.3
REF: p. 327
OBJ: 11.1.3
REF: p. 327
OBJ: 11.1.3
REF: p. 327
OBJ: 11.1.3
3O DIF: L3
3ZnSO
2H O
H
DIF: L3 4Au
2NH
3H O
DIF: L3 Ba(OH)
3H
3Na SO DIF: L3
Zn (PO ) REF: p. 327
OBJ: 11.1.3
2Al
3Cl
2AlCl
PTS: 1 19. ANS: CH
DIF: L3
2O
CO
PTS: 1 20. ANS: Fe (SO ) PTS: 1 21. ANS: 2C H 7O
2Fe(OH)
DIF: L3 6CO
OBJ: 11.1.3 | 11.2.2
REF: p. 327
OBJ: 11.1.3 | 11.2.2
2H O
DIF: L3 3Ba(OH)
REF: p. 327
3BaSO REF: p. 327
OBJ: 11.1.3 | 11.2.2
8H O (incomplete)
PTS: 1 DIF: L3 REF: p. 327 | p. 336 OBJ: 11.1.3 | 11.2.2 22. ANS: C H OH 3O 2CO 3H O (complete) PTS: 1 DIF: L3 OBJ: 11.1.3 | 11.2.2 23. ANS: H (aq) OH (aq) H O(l)
REF: p. 327 | p. 336
PTS: 1 OBJ: 11.3.1 24. ANS: 2K PO 3BaCl
DIF: L3
REF: p. 342 | p. 343
PTS: 1 25. ANS: 0.520 atm
DIF: L2
PTS: 1 26. ANS: 622 mm Hg
Ba (PO ) s
6KCl REF: p. 327
OBJ: 11.3.2
760 mm Hg / 1 atm = 395 mm Hg DIF: L2
REF: p. 387
OBJ: 13.1.2
1 atm / 760 mm Hg = 0.818 atm
PTS: 1 27. ANS: 0K –273.15 C
DIF: L2
REF: p. 387
OBJ: 13.1.2
PTS: 1 28. ANS: 5.85 kPa
DIF: L1
REF: p. 389
OBJ: 13.1.3
PTS: 1 29. ANS:
DIF: L3
REF: p. 392
OBJ: 13.2.3
101.3 kPa
1 atm/101.3 kPa = 1.00 atm
PTS: 1 OBJ: 13.2.4 30. ANS: 90
DIF: L2
REF: p. 387 | p. 395
PTS: 1 31. ANS:
DIF: L2
REF: p. 397
V =V
= 250 mL
PTS: 1 32. ANS:
= 30.0 L
PTS: 1 DIF: L2 33. ANS: T = –55 C + 273 = 218 K T = 30.0 C + 273 = 303 K V
= 1700 mL
DIF: L2
V =V
=V
= 590 mL
PTS: 1 34. ANS:
DIF: L2
P =P
= 340 kPa
PTS: 1 DIF: L2 35. ANS: P V =P V 210 kPa 15.0 L = 790 kPa V
OBJ: 13.3.2
REF: p. 419
OBJ: 14.2.1
REF: p. 421
OBJ: 14.2.1
= 34.3 L
= 820 mL REF: p. 421
OBJ: 14.2.1
= 140 kPa REF: p. 421
OBJ: 14.2.1
REF: p. 419
OBJ: 14.2.1
=V V = 4.0 L PTS: 1 DIF: L3 36. ANS: 227 C + 273 = 500 K 27 C + 273 = 300 K =
;
=
710 kPa
=P
P = 470 kPa PTS: 1 DIF: L3 37. ANS: T = 35.0 C + 273 = 308 K T = 0.0 C + 273 = 273 K V =P
140 mL
= 120 mL
PTS: 1 DIF: L3 38. ANS: T = 27 C + 273 = 300 K; P = 101 kPa T = –10 C + 273 = 263 K; P = 95 kPa V =P
V
V = 3.26
= (101 kP)
DIF: L3
250 mL =
DIF: L2 =
REF: p. 424
OBJ: 14.2.2
REF: p. 427
OBJ: 14.3.1
= 3.0 mol DIF: L2
32 g O
PTS: 1
10 m )
= 0.030 mol
PTS: 1 41. ANS:
PTS: 1 42. ANS: =
(3.5
OBJ: 14.2.2
= 0.25 L
PTS: 1 40. ANS:
P=
REF: p. 424
10 m
PTS: 1 39. ANS:
n=
OBJ: 14.2.1
V
V = 97 kPa
n=P
REF: p. 421
REF: p. 427
OBJ: 14.3.1
= 1 mol O =
= 110 kPa DIF: L2 –(
+
REF: p. 427
OBJ: 14.3.1
) = 95 kPa – (48 kPa + 24 kP) = 23 kPa
DIF: L2
REF: p. 434
OBJ: 14.4.1
43. ANS: = 1.4 PTS: 1 44. ANS: 105
DIF: L2
REF: p. 436
OBJ: 14.4.2
PTS: 1 45. ANS: 4 C
DIF: L2
REF: p. 446
OBJ: 15.1.1
PTS: 1 46. ANS: Molar mass CoCl
DIF: L2
REF: p. 448
OBJ: 15.1.2
6H O = 237.9 g
percentage of water =
100% = 45%
PTS: 1 DIF: L2 47. ANS: 29.8 g = 1.66 mol molar mass 5 = 249.6 g mol CuSO 5H O 5 mol
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