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4) How many grams are there in 7.40 moles of AgNO37
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4) How many grams are there in 7.40 moles of AgNO37

User Crig
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1 Answer

5 votes

Answer:

1258 grams of AgN03

Step-by-step explanation:

We calculate the weight of 1 mol of AgN03:

Weight 1 mol AgN03= Weight Ag + Weight N +( Weight 0)x3=108g+ 14g+16gx3=170 g/mol

1 mol----170 g AgN03

7,4mol---x= (7,4 mol x170 g AgN03)/1 mol=1258 g AgN03

User Skbrhmn
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