34 Ex03 Review

34 Ex03 Review - The cell reaction for a Ni/Cd battery can...

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1 1 The cell reaction for a Ni/Cd battery can be written as 2Ni(OH) 3 (s) + Cd(s) 2Ni(OH) 2 (s) + Cd(OH) 2 (s) Which statements about a Ni/Cd battery is correct? I) Ni is oxidized, and Cd is reduced. II) Cd is oxidized, and Ni is reduced. III) Ni(OH) 3 is the oxidizing agent. IV) Cd is the oxidizing agent. A) I an III B) II and IV C) I and IV D) II and III 2 Electrical Units You need to know these! Electrical potential: Volt, V The voltage supplied to your house is 120 V AC. Electrical charge: Coulomb, C e - = 1.6 x 10 -19 C or 96,485 C/mol Current: Ampere, 1 A = 1 C/s Typical house circuits are rated for 20 A Energy: Joule, 1 J = 1 C x 1 V Power: Watt, 1 W = 1 J/s = 1 A x 1 V Common hair dryers use around 1500 W of power And draw close to 15 A of current
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2 3 The local lighting company charges $0.08 per kilowatt hour. It is charging for A) Voltage provided B) Electrical charge delivered (i.e. number of electrons) C) Current used D) Power used E) Energy used 4 Electrolysis and Battery Discharging and Charging How many grams of Pb(s) are converted into PbSO 4 (s) in the battery if you run the starter motor on your car for 30 s. Assume the motor draws 250 amps of current. Pb 207 g/mol F = 96,485 C/mol. Respond: x.x The following reaction occurs in an automobile battery. PbO 2 (s)+Pb(s)+2H 3 O + (aq)+2HSO 4 - (aq) 2PbSO 4 (s)+4H 2 O(l) mol e- = (250 C/s)(30 s)/96,485 C/mol = 0.0777 mol e- 1 mol Pb produces 2 mol e - so moles Pb = ½ moles e - mass Pb = ½(0.0777 mol e-) x 207 g/mol = 8.04 g Battery provides electrons to run the starter motor. Electrons are produced by converting Pb to PbSO 4 . Determine number of electrons produced in 30 s. Determine amount of Pb converted with those electrons.
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3 5 Given K sp , determine the solubility of a salt. How many grams of lead(II) bromide will dissolve to make exactly 1 L of a saturated solution? PbBr 2 (367 g/mol, K sp = 6.30 x 10 -6 ) 2+ - 2 2+ - 2 2 sp 3- 6 -2 -2 PbBr (s) Pb (aq)+2Br (aq) lots 0 0 -x +x +2x at eq: x mol/L 2 x mol/L K = [Pb ][Br ] = (x)(2x) 4x = 6.30 10 x = 1.16 10 M (1.16x10 mol /L)(1 L)(367 g/mol) = 4.27 g   x x The solubility of lead(II) bromide is 4.27 g/L.
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This note was uploaded on 02/13/2011 for the course CHE 132 taught by Professor Hanson during the Spring '08 term at SUNY Stony Brook.

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34 Ex03 Review - The cell reaction for a Ni/Cd battery can...

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