Practice Exam I Fall 2004 -- ANSWERS updated

Practice Exam I Fall 2004 -- ANSWERS updated - CHEM 105...

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CHEM 105 – Exam I Fall 2003 1) Determine the number of protons, neutrons, and electrons in each of the following. Also, provide the atomic symbol and name if so asked. ( 13 pts total) The neutral alkali metal in the fourth row of the periodic table and with a mass number of 41. P ____19_______ N ____22_______ E _____19______ Atomic Symbol ______K_________ Atomic Name _____Potassium__________ The ion of 108 Ag (note you must know the charge) P _____47______ N _____61______ E ______46_____ Atomic Name ____Silver___________ A neutral tin-121 atom P ___50________ N ___71________ E ____50_______ Atomic Symbol _____Sn__________ 1
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2) Provide the name or formula, as appropriate for the following: (14 pts total) Pb(HPO 4 ) 2 Lead (IV) hydrogen phosphate Manganese (II) oxide MnO NH 4 NO 2 Ammonium Nitrite HNO 2 (aq) nitrous acid NO 3 Nitrogen trioxide Gallium chlorite Ga(ClO 2 ) 3 Tin (IV) sulfate Sn(SO 4 ) 2 3) Write a balanced chemical equation (including physical states) for the reaction of aqueous potassium carbonate with liquid hydrochloric acid to form gaseous CO 2 , liquid water, and aqueous potassium chloride. (8 pts total) K 2 CO 3 (aq) + 2HCl(l) CO 2 (g) + H 2 O (l) + 2KCl(aq) 2
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4) How many grams of C 4 H 10 must be burned (that is, combusted) to form 22.45 L of CO 2 at 27°C and 750 torr. (10 pts) C 4 H 10 + 13/2 O 2 4CO 2 (g) + 5 H 2 O(l) n = PV/RT = (750/760)(22.45)/[(0.0821)(300)] = 0.8995 moles CO 2 0.8995 moles CO 2 x 1 mole C 4 H 10 x 58 grams C 4 H 10 = 13.0 g 4 mole CO 2 x 1 mole C 4 H 10 5) A 1 mole sample of water vapor at 400 K occupies 32.84 L at 1 atm. Calculate the number of mL occupied at 200 K. Knowing that the density of water at 200 K is about 1g/mL compare the density of the water vapor at 200 K from your calculated volume. Comment on why this result is so far off. (10 pts) V 1 /T 1 = V 2 /T 2 32.84/400 = V 2 /200 16.42 L D = g/mL = 18g/16420 mL = 0.0011 g/mL which is much too low because at 200 K water is a liquid NOT a gas so the ideal gas law breaks down. 3
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6) Given the BALANCED chemical equations below, calculate the actual yield of Mn 3 (PO 4 ) 2 of the reaction of 1.2 grams of H 2 C 2 O 4 with 0.7 grams of KMnO 4 if the percent yield of the first step is 95% and the percent yield of the second step was 92%. Assume that all chemicals other than H 2 C 2 O 4 and KMnO 4 are present in large excess. (15 pts) 5H 2 C 2 O 4 + 2KMnO 4 + 6H + 10CO 2 + 2Mn +2 8H 2 O + 2K + (step 1: 95% yield) 3Mn +2 + 2 PO 4 -3 Mn 3 (PO 4 ) 2 (Step 2: 92% yield) 1.2 g H 2 C 2 O 4 x 1 mole H 2 C 2 O 4 x 2 mole Mn +2 = 0.00533 moles Mn +2 90 g H 2 C 2 O 4 x 5 mole H 2 C 2 O 4 0.7 g KMnO 4 x 1 mole KMnO 4 x 2 mole Mn +2 = 0.00443 moles of Mn +2 (limiting) 158.03 g KMnO 4 x 2 mole KMnO 4 0.00443 mole Mn +2 x 1 mole Mn 3 (PO 4 ) 2 x 354.8 g Mn 3 (PO 4 ) 2 = 0.523 g 3 mole Mn +2 x 1 mole Mn 3 (PO 4 ) 2 0.523 g x 0.95 x 0.92 = 0.457 g 0.5 g (1 SF due to the 0.7 g ) 4
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7) What is the density of and v rms or a gas that effuses four times slower than ammonia? The conditions are 1 atm and 300K.
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This test prep was uploaded on 04/20/2008 for the course CHEM 105 taught by Professor F during the Fall '05 term at Grove City.

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Practice Exam I Fall 2004 -- ANSWERS updated - CHEM 105...

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