M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?
m(Mg)=wm
m(Zn)=(1-w)m
Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)
Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)
m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)
w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))
w=0.583 (58.3%)
If you overheat copper sulfate higher of mass will be lost that is copper sulfate will loss sulfur and oxygen which led to a higher loss of mass than if you would have heated enough. This higher mass lost will be shown in calculation as percentage of water lost
Answer:
The answer to your question is 175 gallons
Explanation:
Data
60% antifreeze solution volume = ? = x
25% antifreeze solution volume = 25%
Final concentration 50%
Process
0.6x + 0.25(70) = 0.5(70 + x)
Simplification
0.6x + 17.5 = 35 + 0.5x
0.6x - 0.5x = 35 - 17.5
0.1x = 17.5
x = 17.5/0.1
Result
x = 175 gallons
The answer is d I think. Don’t quote me if I’m wrong
