Question #189219

How many grams is 5.97E24 molecules of C6H12O6 gas at STP?


1
Expert's answer
2021-05-12T06:26:26-0400
m/M=N/Nam/M=N/Na

m=MN/Nam=MN/Na

m=180∗5.97E24/6.02E23=1785gm=180*5.97E24/6.02E23=1785g


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