PExamNew3A - Chemistry 101 Sample Exam #3 Note: This exam...

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Unformatted text preview: Chemistry 101 Sample Exam #3 Note: This exam has more than 100 points; your exa m w ill be 100 po ints, only . Multiple Choice. There is only on e “b est” answ er for each of these multip le- choice questions. Please note, all answer s must be given to th e correct numb er of significan t f igures! (4 points each) 1. A 4.09 g sample of a pure substance has a volume of 5.98 mL. The volume of 50.0 grams of this substance is: a) 73.105 mL b) 7.3105 x102 mL *c) 73.1 cm3 d) Insufficient information is provided to correctly calculate this answer. e) None of the above is correct. 2. When 13.99 grams of a substance burns, 24.06 grams of H2O are produced. When 1.2×103 grams of this same substance is burned, the amount of water produced will be: *a) 2.1 x 103 grams b) 2,063.7 grams c) 2,063.76 grams d) Insufficient information is provided to correctly calculate this answer. e) None of the above is correct. 3. A sample of drink ing water contains 988 ppm (par ts per million) As. In 82.0 grams of that same w ater: *a) b) c) d) e) The A s con centr ation w ill b e 988 ppm. There will be 988.0 grams of As. There will be 82.0 ppm of As. Insufficient information is provided to correctly calculate this answ er. None of the above is co rrect. 4. A compound contains 24.96 % C, by mass. In 500.0 g of the compound there will be: a) 12,480 grams of carbon. b) 1.2 x 102 grams of carbon. *c) 124.8 grams of carbon. d) Insufficient information is provided to correctly calculate this answer. e) None of the above is correct. 5. 125.0 µL is equivalent to: a) 1.250 x 10-6 L b) 1.250 mL *c) 1.250 x 10-1 cm3 d) Both a and c are correct. e) None of the above is correct. 6. 6.02×1023 kilograms is equivalent to: a) 6.02×1026 grams b) 6.02×1020 grams c) A mole of kilograms. *d) Both a and c are correct. e) None of the above is correct. H C N O 1.0079 12.011 14.00674 15.9994 F Na Cl Ca 18.9984032 22.98977 35.4527 40.078 7. Using the data given in the table above, the formula mass of methane, CH4, is: a) 16.0426 g/mol *b) 16.043 g/mol c) 13.0189 g/mol d) 13.019 g/mol e) None of the above is correct. 8. The molar mass of KHC8H4O4 is 204.224 g/mol. A sample containing 0.0100 mole of KHC8H4O4 will: a) have a mass of 204.224 grams. b) have a density of 2.04 g/cm3 *c) contain 6.02 x 1021 formula units. d) Both a and c are correct. e) None of the above is correct. 9. Which of the following statements is true regarding the reaction given below: 2 Ta + 5 Br2 2 TaBr5 a) The equation is properly balanced. b) One mole of Ta will yield one mole of TaBr5. c) The oxidation state of Ta changes from 0 to +5 in this reaction. d) Both a and c are correct. *e) a, b, and c are correct. 10. In 1.50 moles KHC8H4O4 there are: a) 1.50 moles of potassium. b) 1.50 moles of hydrogen. c) 6.00 moles of oxygen. *d) Both a and c are correct. e) None of the above is correct. 11. If a mole contains 6.02 × 1023 molecules, 10% of a mole contains: a) 6.02 × 1020 molecules b) 1.00 × 1022 molecules c) 0.602 × 1022 molecules d) Both a and c are correct. *e) None of the above is correct. 12. With regard to the reaction of sodium with chlorine (2 Na + Cl2 2 NaCl); if 1.00 mole of sodium is heated in the presence of an excess amount of chlorine, how many moles of NaCl will be formed? a) 6.02 × 1023 formula units b) 2.00 moles c) 1.00 mole *d) Both a and c are correct. e) None of the above is correct. 13. One liter of a solution is found to contain 15.25 grams of NaCl. How many liters of this same solution must you have to in order to have a total of one mole of NaCl molecules? *a) 3.832 liters b) 0.2609 liters c) 0.261 liters d) Insufficient information is provided to correctly calculate this answer. e) None of the above is correct. 14. If 6.02 × 1020 molecules of carbon burn in air (C + O2 be produced? Answer should be to 2 sig figs CO2), how many grams of CO2 will a) 4.4 × 1020 grams of CO2 b) 0.44 grams of CO2 *c) 4.4 × 10-2 grams of CO2 d) Insufficient information is provided to correctly calculate this answer. e) None of the above is correct. 15. A beaker contains 0.23 cm3 of mercury. If an additional 4.0 mL of mercury is added to this beaker, the total mass of mercury in the beaker will be: a) 4.25 grams. b) 4.250 grams. c) 4.2 grams. *d) Insufficient information is provided to correctly calculate this answer. e) None of the above is correct. Use the equation below to answer questions 16-19: C3H8 + 5 O2 3 CO2 + 4 H2O 16. In order to produce 25 .0 moles of CO2, how many moles of molecular oxygen are required? a) 75.0 moles b) 83.2 moles *c) 41.7 moles d) Insufficient information is provided to correctly calculate this answer. e) None of the above is correct. 17. If 2.00 moles of C3H8 react, how much CO2 will be formed? a) 3.00 moles *b) 6.00 moles c) 2,063.76 grams d) Both a and c are correct. e) None of the above is correct. 18. If 12.0 moles of water are formed in the reaction, how many moles of C3H8 were consumed? *a) 3.00 moles b) 4.00 moles c) 12.0 moles d) Insufficient information is provided to correctly calculate this answer. e) None of the above is correct. 19. The combustion of 1.00 mole of C3H8 is found to produce 68 .3 grams of w ater vapor. The p ercen tag e yield in th is r eactio n is: a) b) *c) d) e) 68.3% 47.4% 94.8% This result canno t be calculated w ithout knowing th e volume of the water vapor. None of the above is co rrect. 20. Which of the following is tru e reg arding a reaction in which one reactant is present in stoichiometric excess over ano ther reactan t? a) *b) c) d) e) The product which is formed in excess of th e theoretical y ield is termed “excess p roduct”. The numb er of mo les of product is deter min ed by th e numb er of moles of the lim iting reagent. The p ercen tag e yield of th e r eaction will be reduced. Both a and c are correct. None of th e above is corrrrect 21. Pyrite (FeS2), presen t in coal, is responsible for sulfuric acid H2SO4 mine drainage. How many grams of sulfuric acid are produced from 1.0 g of FeS2? (Ato mic mass: Fe = 55 .85, O = 16.00, H = 1 .008, S = 32.06). 2 FeS2 + 2 H2O + 7 O2 a) *b) c) d) e) 2 FeSO4 + 2 H2SO4 1.23 0.82 0.41 1.6 None of the above. 22. Which of the fo llow ing co mpounds are likely to be insoluble (or very slightly so luble) in water?: a) b) c) *d) e) BaSO4 Pb I 2 CuNO3 Both a and b will be inso luble. All of th e above compounds will be sign ificantly so luble in water. 23. Barium ch loride reacts w ith potassium carbonate the precipitate is *a) b) c) d) e) barium carbonate potassium chlor ide barium carbonate and po tassium ch loride There is no precip itate formed None of the above is co rrect 24. The reaction shown b elow : H2 + Br2 a) b) c) *d) e) 2 H Br can b e describ ed as a synthesis reaction. can b e describ ed as a single-displa cem ent reaction. can b e describ ed as an oxidation-reduction r eaction. Both a and c are correct. a, b, and c are correct. 25. 2Al + 3S Al2S3 In the reaction g iven above, th e oxid ation number of Al: a) changes from 0 to +4. b) does not change. *c) is 0 in the reactants. d) Both a and c are correct. e) None of the above is correct. Calcula tions. In the space provided, clearly set up the requ ired calcu lations and repor t the answer s to th e appropriate nu mber of significan t f igures. Fu ll cr edit requires th at all calculation and ratios be shown. (8 points ea ch) H C N O 1.0079 12.011 14.00674 15.9994 F Na Cl Ca 18.9984032 22.98977 35.4527 40.078 21. Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 % N. b. If the molar mass is around 186g what is the molecular formula? 38.67 g C x 1mol C 12.01 g C = 3.220 mol C The ratio is 3.220 mol C 3.219 mol N 16.22 g H x 1mol H = 16.09 mol H The ratio is 16.09 mol H 1.008 g H 45.11 g N x 1mol N = 1 mol C/N = 5 mol H/N 3.219 mol N = 3.219 mol N So the empirical formula is CH5N 14.01 g N The mass of CH5N is 12.01 + 5(1.008) + 14.01 = 31.06 g. The molar mass is 186g / 31.06 = 6 so molecular formula is C6H30N6 22. Calculate the mass percentage of carbon, in ethanol, CH3CH2OH. Report your answer using the correct number of SF. Molar mass: 2(12.011) + 15.9994 + 6(1.0079) = 46.069 " 12.011 % 2$ ' ( 100 = 52.144% # 46.069 & 23. Calculate the number of moles of CCl4 in 25 mL of CCl4. The density of CCl4 is 1.45 g cm-3. Report your answer using the correct number of SF. Molar mass: 12.011 + 4(35.4527) = 153.822 % ' cm & " % = 0.24 moles $153.822 g mole ' # & (25 " cm 3 )$1.45 g # 3 24. Determine the number of moles of CaCl2 that contains 0.012 grams of Ca. Report your answer using the correct number of SF. " 1 mol Ca % )4 $ ' " % $ 1 mol CaCl2 ' = 3.0 ( 10 moles & $ 40.078 g mole' # # & (0.012 g) 25. A sample contains 12.1 grams of H2O; calculate the number of moles of H which are in this sample. Report your answer using the correct number of SF. Molar mass: 15.9994 + 2(1.0079) = 18.0152 " 2 mol H % $ ' = 1.34 moles ' " %$ $18.0152 g mole' # 1 mol H 2O & # & (12.1 g) 26. A sample of nitric acid, HNO3, has a mass of 4.062 g, calculate the number of moles of nitric acid this represents. Report your answer in scientific notation using the correct number of SF. Molar mass: 1.0079 + 14.00674 + 3(15.9994) = 63.0128 (4.062 g) )2 " % = 6.446 (10 moles $63.0128 g mole' # & 27. P4O10 reacts with water to form phosphoric acid according to the (unbalanced) equation given below: P4O10 + H2O H3PO4 a. Write a balanced equation for this reaction (4 points). P4O10 + 6 H2O 4 H3PO4 b. Calculate the maximum mass of H3PO4 that can be produced from a mixture of 10.00 g P4O10 and 2.00 g H2O. (Molar masses: P4O10 = 283.886 g mol-1, H2O = 18.0148 g mol-1, H3PO4 = 97.9937 g mol1 ) (8 points) Simply calculate the mass that each reactant would yield: " 4 moles H PO %" 97.9937 g H PO % 3 4 '$ 3 4' $ " % $ mole P4 O1 0 '$ 1 mole H 3PO 4 ' = 13.81 g H3 PO4 &# & $283.886 g P4 O1 0 mole' # # & (10.00 g P4 O1 0) " 4 moles H PO %" 97.9937 g H PO % 3 4 '$ 3 4' $ " % $ 6 moles H2 O '$ 1 mole H3 PO4 ' = 7.25 g H3 PO4 g H2 O # &# & $18.0148 mole' # & (2.00 g H2 O) P4O10 c. Which is the reactant in excess? ________________________ (4 pts) ...
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This note was uploaded on 04/11/2010 for the course CHEM 101 taught by Professor Denofrio during the Spring '08 term at University of Illinois at Urbana–Champaign.

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