Question

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

Answer 1

`The\text{ number of atoms of Ag in AgNO}_3\text{ is 2.983}*\text{ 10}^(23)\text{ atoms}`

Step-by-step 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 = }(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 = }(x)/(6.022*10^(23)) \\ \text{ Cross multiply} \\ \text{ x = 0.4954 }*\text{ 6.022}*10^(23) \\ \text{ x = 2.983 }*\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 }*\text{ 2.983 }*\text{ 10}^(23) \\ \text{ Number of atoms of Ag = 2.983 }*\text{ 10}^(23)\text{ atoms} \end{gathered}`