calculate molecular present in 42 gram of sodium hydrogen carbonate NaHCO3 Na=23, H=1, C=12, O=16
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we will first find its molecular mass
23+1+12+3*16=84
42/84=0.5
means 0.5 moles
so moles *Na =no. of molecules
0.5*6.022*10^23
=3.011*10^23
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23+1+12+3*16=84
42/84=0.5
means 0.5 moles
so moles *Na =no. of molecules
0.5*6.022*10^23
=3.011*10^23
pls mark branliest if eligible
Answered by
2
its no. of molecules = 3.011×10^23
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