Asked by Haley
An excess of zinc metal is added to 50.0 mL of a 0.100 M AgNO3 solution in a constant-pressure calorimeter. Assuming that the density of the solution is the same as for water, calculate the mass of the solution.
Answers
Answered by
DrBob222
The question doesn't make sense to me. The 50 mL of the AgNO3 solution has a mass of 50.0 g assuming the density of water and the AgNO3 soln is the same; however, that is the mass before an excess of Zn metal is added. I have no idea how much the total material weighs. It all depends upon the amount of Zn added.
Answered by
Chris
50mL of 0.1M = .005 moles of AgNO3, so based on balanced equation, that will react with .0025 moles Zn.
Calculate q(zinc) = -(q(sol) + q(cal))
q(zinc) = -(50*4.18*2.92 + 98.6*2.92)
q(zinc) = -898.19 J
then divide by .0025 moles of Zn reacting (and covert to kJ)
= -306kJ/mol
Calculate q(zinc) = -(q(sol) + q(cal))
q(zinc) = -(50*4.18*2.92 + 98.6*2.92)
q(zinc) = -898.19 J
then divide by .0025 moles of Zn reacting (and covert to kJ)
= -306kJ/mol
Answered by
Chris
50mL of 0.1M = .005 moles of AgNO3, so based on balanced equation, that will react with .0025 moles Zn.
Calculate q(zinc) = -(q(sol) + q(cal))
q(zinc) = -(50*4.18*2.92 + 98.6*2.92)
q(zinc) = -898.19 J
then divide by .0025 moles of Zn reacting (and covert to kJ)
= -360kJ/mol
Calculate q(zinc) = -(q(sol) + q(cal))
q(zinc) = -(50*4.18*2.92 + 98.6*2.92)
q(zinc) = -898.19 J
then divide by .0025 moles of Zn reacting (and covert to kJ)
= -360kJ/mol