Question

Pls help me I don’t know how to do this

Answer 1

ANSWER

$The\text{ number of atoms of Ag in AgNO}_3\text{ is 2.983}\times\text{ 10}^{23}\text{ atoms}$

EXPLANATION

Given information

The mass of silver nitrate (AgNO3) = 84.15 grams

Avogadro's constant = 6.022 x 10^23

To find the number of atoms of AgNO3, follow the steps below

Step 1: Find the number of moles using the below formula

$\text{ Mole = }\frac{\text{ mass}}{molar\text{ mass}}$

Recall, that the molar mass of AgNO3 is 169.87 g/mol

Step 2: Substitute the given data into the formula in step 1

$\begin{gathered} \text{ Mole = }\frac{84.15}{169.87} \\ \text{ Mole = 0.4954 mol} \end{gathered}$

Hence, the number of moles of AgNO3 is 0.4954 mole

Step 3: Find the number of atoms of AgNO3 using the formula below

$\begin{gathered} \text{ mole = }\frac{\text{ number of atoms}}{Avogadro^{^{\prime}}s\text{ number}} \\ \end{gathered}$

Let the number of atoms be x

$\begin{gathered} \text{ 0.4954 = }\frac{x}{6.022\times10^{23}} \\ \text{ Cross multiply} \\ \text{ x = 0.4954 }\times\text{ 6.022}\times10^{23} \\ \text{ x = 2.983 }\times\text{ 10}^{23}\text{ atoms} \end{gathered}$

Since we have 1 atom of Ag in AgNO3, hence, the number of atoms of Ag in AgNO3 is calculated below

$\begin{gathered} \text{ Number of atoms of Ag = 1 }\times\text{ 2.983 }\times\text{ 10}^{23} \\ \text{ Number of atoms of Ag = 2.983 }\times\text{ 10}^{23}\text{ atoms} \end{gathered}$