Chemistry, asked by hetviviradiya49831, 1 year ago

The no. Of Br- ions in 2.1 gm CaBr2 is

Answers

Answered by nicolemakara
0

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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