All categories

3 years ago

Calculate the number of ions present in 8.2 moles of calcium nitrate

Chemistry

1 answer:

Tamiku [17]3 years ago

5 0

Answer:

1.48 × 10²⁵ Ions

Solution:

Calcium Nitrate having a chemical formula Ca(NO₃)₂ is made up of one Ca²⁺ cation and two NO₃⁻ anions. Hence, we can say that each Calcium Nitrate contains three ions respectively.

Therefore,

1 mole of Ca(NO₃)₂ contains = 6.022 × 10²³ formula Units

So,

8.2 mole of Ca(NO₃)₂ will contain = X Formula Units

Solving for X,

X = (8.2 mol × 6.022 × 10²³ formula units) ÷ 1 mol

X = 4.93 × 10²⁴ Formula Units

As discussed,

1 Formula Unit of Ca(NO₃)₂ contains = 3 Ions

So,

4.93 × 10²⁴ Formula units will contain = X ions

Solving for X,

X = (4.93 × 10²⁴ Formula Unit × 3 Ions) ÷ 1 Formula Unit

X = 1.48 × 10²⁵ Ions

You might be interested in

Write the balanced equation and draw the particulate representation of the following chemical reaction: sodium metal reacts with

KonstantinChe [14]

Answer:

2Na + Cl2 ------> 2NaCl

4H2O2 --------> 4H2O + 2O2

CH4 + 2O2 -------> CO2 + 2H2O

Explanation:

5 0

3 years ago

Given the following two quantities: 0.50 mol of CH4 and 1.0 mol of HCl,

Vinil7 [7]

Answer:

(a) HCl

(b) HCl

(d) HCl

Explanation:

<em>Given: </em>0.50 mol of CH₄ and 1.0 mol of HCl

Using stoichiometry we can calculate the answers to parts a, b, c, and d.

# of moles × Avogadro's number = # of atoms or molecules

Avogadro's number: 6.02 * 10²³

$\displaystyle 0.50\ \text{mol CH}_4 \cdot \frac{6.02\cdot 10^2^3 \ \text{atoms CH}_4}{1 \ \text{mol CH}_4} = 3.01 \cdot 10^2^3 \ \text{atoms CH}_4$
$\displaystyle 1.0\ \text{mol HCl} \cdot \frac{6.02\cdot 10^2^3 \ \text{atoms HCl}}{1 \ \text{mol HCl}} = 6.02 \cdot 10^2^3 \ \text{atoms HCl}$

HCl has more atoms than CH₄.

This is calculated the same way as Part (a); HCl has more molecules than CH₄.

Molar mass of CH₄ = 16.04 g/mol

Molar mass of HCl = 36.458 g/mol

$\displaystyle 0.50\ \text{mol CH}_4 \cdot \frac{16.04 \ \text{g CH}_4}{1 \ \text{mol CH}_4} = 8.02 \ \text{g CH}_4$
$\displaystyle 1.0\ \text{mol HCl} \cdot \frac{36.458 \ \text{g HCl}}{1 \ \text{mol HCl}} = 36.458 \ \text{g HCl}$

HCl has a greater mass than CH₄.

Assuming STP:

Molar volume of any gas at STP is 22.4 L/mol.

$\displaystyle 0.50\ \text{mol CH}_4 \cdot \frac{22.4 \ \text{L CH}_4}{1 \ \text{mol CH}_4} = 11.2 \ \text{L CH}_4$
$\displaystyle 1.0\ \text{mol HCl} \cdot \frac{22.4 \ \text{L HCl}}{1 \ \text{mol HCl}} = 22.4 \ \text{L HCl}$