2009Fchem125exam2answers040ri66

2009Fchem125exam2answers040ri66 - Chem. 125 Exam #2 Fall...

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Chem. 125 Exam #2 Fall 2009 PART A. (4 Points each) 1. What is the value of a 0.50 in 3 of gold, if a troy ounce is worth $1,000.00? Useful information: density of Au = 19.3 g/cm 3 ; 1 troy ounce (t oz) = 31.1 g; 2.54 cm = 1 in. a. <$1,000 b. Between $1,000 and $2,000 c. Between $2,000 and $4,000 d. Between $4,000 and $5,000 e. >$5,000 2. Which of the following ionic compounds is soluble in water? a. MgCO 3 b. Iron (II) phosphate c. Iron (III) sulfide d. HgI e. KOH 3. What precipitate is formed when a solution of silver acetate is mixed with a solution of zinc iodide? a. There is no precipitate b. ICH 3 COO c. ZnAg d. AgI e. Zn(CH 3 COO) 2 4. 3.50 g ammonium sulfide, (NH 4 ) 2 S, is dissolved in enough water to make 500. mL of solution. What is the concentration of the ammonium ions? a. 0.102 M b. 0.205 M c. 0.051 M d. 0.306 M e. > 1.0 M 3 3 3 0.5 2.54 19.3 1 $1000 $5,084 $5,100 1 31.1 1 in cm g t oz x x x x in cm g t oz = ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 4 2 4 4 2 2 4 2 4 14.01 1.01 4 2 32.07 68.3 / 3.50 1 / 0.500 0.102 / 0.102 1 68.3 2 0.1025 0.205 MM NH S x x g mol g NH S mol NH S x L mol L M NH S g NH x M + = + + = = = = =
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Chem. 125 Exam #2 Fall 2009 5. The stockroom has 0.650 M H 2 SO 4 . How many milliliters of this acid are required to prepare 250. mL of 0.0425 M sulfuric acid? a. 16.3 mL b. 65.4 mL c. 42.0 mL d. 27.3 mL e. 3.82 mL 6. Approximately how many potassium ions are in 20 mL of 0.44 M potassium phosphate, K 3 PO 4 ? a. 2 x 10 22 b. 5 x 10 21 c. 8 x 10 23 d. 3 x 10 23 e. 50 7. What is the concentration of nitric acid if excess zinc powder is added to75.0 mL of the acid to produce 0.13 moles of hydrogen gas? ( 29 ( 29 ( 29 ( 29 ( 29 3 3 2 2 2 . Zn s HNO aq Zn NO aq H g + + a. 1.7 M b. 0.26 M c. 3.5 M d. 0.47 e. 5.2 M 8. What is the temperature of an ideal gas where P = 700. torr, n = 0.65 mol, and V = 6.00 L? a. 104 K b. 267 K c. 43.9 K d. 67.6 K e. 78,700 K 2 4 2 4 0.0425 0.25 1000 16.3 1 0.65 mol H SO L L mL x x x mL L mol H SO L = 3 2 3 2 3 3 3 2 0.13 0.26 1 1 0.26 3.47 3.5 0.075 mol HNO mol H x mol HNO mol H mol HNO mol Conentration HNO M HNO L L = = = ( 29( 29 ( 29 700 1 0.921 1 760 0.921 6.00 ; 104 0.65 0.08206 torr atm P x atm torr atm L PV PV nRT T K L atm nR mol mol K = = = = = = 23
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This note was uploaded on 10/28/2010 for the course CHEM 005 taught by Professor Ellis during the Fall '08 term at NJIT.

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2009Fchem125exam2answers040ri66 - Chem. 125 Exam #2 Fall...

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