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Revised 6/22/17 Station # Date me- - -- CHM 112 Vapor Pressure and Heat of Vaporization...
Revised 7/25/19 Name Station Date CHM 112 Determination of an Equivalent Mass by Electrolysis Pre Lab...
Revised 7/25/19 Name Station Date CHM 112 Determination of an Equivalent Mass by Electrolysis Pre Lab Questions 1- In the following reactions determine a: which element is oxidized, b: which element is reduced, c: oxidizing agent, d: reducing agent: A- Fe2+ + H+ + Cr2O7 Felt + C + H2O B- Fe2S3 + 12HNO3 - 2Fe(NO3)3 + 35 + 6NO2 + 6H20 C-1 + MnO +H* - 12 + MnO2 + H20 2- Balance the following equations: A- MnO4 +...
Revised 7/25/19 Station CHM 112 Determination of an Equivalent Mass by Electrolysis Post Lab Ouestions 1....
Revised 7/25/19 Station CHM 112 Determination of an Equivalent Mass by Electrolysis Post Lab Ouestions 1. In an electrolysis cell similar to the one employed in this experimenta student observed that his unknown metal anode lost 0.20 while a total volume of 9630 ml of was being produced. The temperature in the laboratory was 25°C and the barometric pressure was 748 mm Hg. At 25°C the vapor pressure of water is 23.8 mm Hg. To find the equivalent mass of...
Revised 7/25/19 Name Station Date CHM 112 Determination of an Equivalent Mass by Electrolysis Pre Lab Questions 1- In the following reactions determine a: which element is oxidized, b: which element is reduced, c: oxidizing agent, d: reducing agent: A- Fe2+ + H+ + Cr2O7 Felt + C + H2O B- Fe2S3 + 12HNO3 - 2Fe(NO3)3 + 35 + 6NO2 + 6H20 C-1 + MnO +H* - 12 + MnO2 + H20 2- Balance the following equations: A- MnO4 +...
Revised 7/25/19 Station CHM 112 Determination of an Equivalent Mass by Electrolysis Post Lab Ouestions 1. In an electrolysis cell similar to the one employed in this experimenta student observed that his unknown metal anode lost 0.20 while a total volume of 9630 ml of was being produced. The temperature in the laboratory was 25°C and the barometric pressure was 748 mm Hg. At 25°C the vapor pressure of water is 23.8 mm Hg. To find the equivalent mass of...
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Assume Kw = 1.01 ✕ 10−14
For pure water, we can calculate [H3O+ ] = [OH...
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