Question

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

Answer 1

Answer:

1258 grams of AgN03

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