Question

question 9
6 pts
how many moles of silver ions are present in 32.46 g of agno₃?
5520 mol
5.23 mol
0.191 mol
32.5 mol
question 10
6 pts
how many moles are present in 1.64 g of mgso₄?
1.64 mol
0.0136 mol
0.734 mol

1. mol

Explanation:

Step1: Calculate molar mass of AgNO₃

The molar mass of AgNO₃: $M_{AgNO_3}=107.87 + 14.01+3\times16.00 = 169.88\ g/mol$.

Step2: Determine moles of AgNO₃

Using the formula $n=\frac{m}{M}$, where $m = 32.46\ g$ and $M = 169.88\ g/mol$. So $n_{AgNO_3}=\frac{32.46}{169.88}\approx0.191\ mol$. In AgNO₃, there is 1 silver ion per formula - unit, so moles of silver ions is equal to moles of AgNO₃, which is 0.191 mol.

Step3: Calculate molar mass of MgSO₄

The molar mass of MgSO₄: $M_{MgSO_4}=24.31+32.07 + 4\times16.00=120.38\ g/mol$.

Step4: Determine moles of MgSO₄

Using the formula $n=\frac{m}{M}$, where $m = 1.64\ g$ and $M = 120.38\ g/mol$. So $n_{MgSO_4}=\frac{1.64}{120.38}\approx0.0136\ mol$.

Answer:

Question 9: 0.191 mol
Question 10: 0.0136 mol