Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
A student begins with 9.7 mL of a 0.26 M Cu(NO3)2 solution and performs copper cycle lab. She recovers 0.075 grams of copper me...Asked by B
A student begins with 9.7 mL of a 0.26 M Cu(NO3)2 solution and performs copper cycle lab. She recovers 0.075 grams of copper metal.
A) What is the Theoretical Yield
B) What is the Actual Yield
C) What is the Percent Yield
A) What is the Theoretical Yield
B) What is the Actual Yield
C) What is the Percent Yield
Answers
Answered by
DrBob222
I don't know what a copper cycle lab is.
a. mols Cu(NO3)2 = M x L = ?
1 mol Cu in 1 mol Cu(NO3)2
g Cu = mols x atomic mass = theoretical yield
b. 0.075 from the problem.
c. (actual/theoretical)*100 = %yield
a. mols Cu(NO3)2 = M x L = ?
1 mol Cu in 1 mol Cu(NO3)2
g Cu = mols x atomic mass = theoretical yield
b. 0.075 from the problem.
c. (actual/theoretical)*100 = %yield
Answered by
B
Thank you! I was just missing the step of multiplying by atomic mass! :)
There are no AI answers yet. The ability to request AI answers is coming soon!