Question

3.) how many formula units (particles of agno₃) are in 5.50 grams of agno₃? (1 point)
8.67×10²¹ formula units
0.0032 formula units
6.14 and 10²² formula units
1.93×10²² formula units
4.) how many molecules are in 5.00 grams of nh₃? (1 point)
0.333
6.58×10²² molecules
1.77×10²³ molecules
2.00×10²³ molecules

Explanation:

Step1: Calculate molar mass of AgNO₃

The molar mass of Ag (silver) is approximately 107.87 g/mol, N (nitrogen) is 14.01 g/mol, and O (oxygen) is 16.00 g/mol. For AgNO₃, $M = 107.87+14.01 + 3\times16.00=169.88$ g/mol.

Step2: Calculate moles of AgNO₃

Use the formula $n=\frac{m}{M}$, where $m = 5.50$ g and $M = 169.88$ g/mol. So $n=\frac{5.50}{169.88}\approx0.0324$ mol.

Step3: Calculate formula - units of AgNO₃

Use Avogadro's number $N_A = 6.022\times 10^{23}$ formula - units/mol. The number of formula - units $N=n\times N_A=0.0324\times6.022\times 10^{23}\approx1.95\times 10^{22}$ formula - units.

Step1: Calculate molar mass of NH₃

The molar mass of N (nitrogen) is 14.01 g/mol and H (hydrogen) is 1.01 g/mol. For NH₃, $M = 14.01+3\times1.01 = 17.04$ g/mol.

Step2: Calculate moles of NH₃

Use the formula $n=\frac{m}{M}$, where $m = 5.00$ g and $M = 17.04$ g/mol. So $n=\frac{5.00}{17.04}\approx0.293$ mol.

Step3: Calculate number of molecules of NH₃

Use Avogadro's number $N_A = 6.022\times 10^{23}$ molecules/mol. The number of molecules $N=n\times N_A=0.293\times6.022\times 10^{23}\approx1.77\times 10^{23}$ molecules.

Answer:

1.93 X 10²² formula units