what is the question ? I don't see any question but a few words
Answer:
66.4 mL
Explanation:
A 75.7% (v/v) value, means that f<u>or every 100 mL of rubbing alcohol, there are 75.7 mL of isopropanol.</u>
With the above information in mind, we can s<u>olve the problem by multiplying 87.7 mL by 75.7 %</u>:
87.7 mL * 75.7 / 100 = 66.4 mL
So there are 66.4 mL of isopropanol in 88.7 mL of rubbing alcohol.
Answer:
Explanation:
Na2CO3+Ca(NO3)2=CaCO3+2NaNO3
nNa2CO3=0.02
nCa(NO3)2=0.02
mCaCO3=0.02*100=2 gram
nNaNo3=0.04
Cm=2/15
