The no. Of Br- ions in 2.1 gm CaBr2 is
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Answer:
n(CaBr2)=m/M
n=2.1/199
n=0.01mol
n(Br-)=n(CaBr2) ×2
n(Br-)=0.02mol
Explanation:
for every CaBr2, 2Br- is needed... so multiply the n(CaBr2) by 2.
I hope this helps
abuahmed2200a:
Then multiply in avogadro's number which is 6.02×10^23
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