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1) How many grams are there in 2.35x10^26 molecules of aluminum acetate? 2) What volume would the aluminum acetate in Problem 1 occupy at STP?

1) How many grams are there in 2.35x10^26 molecules of aluminum acetate?
2) What volume would the aluminum acetate in Problem 1 occupy at STP?

Grade:9

2 Answers

Vikas TU
14149 Points
7 years ago
Part 1 =>
moles of aluminium acettae = > 2.35 x 10^26/(6.023 x 10^23) = > 0.390 x 10 ^3 mol.
weight = moles x molar mass of Al(CH3COO)3.
           = 0.390 x 10 ^3 x 204
            = 79.56 gm.
 
Part 2 = > 
Volume  = > moles x 22.4 = > 390 x 22.4 = > 8736 litre.
Doniboi
10 Points
7 years ago
Hey why is the moles in part 2 390? Pls explain it to me. It would be awesome if you tell it to me. And thanks for the answer
 

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