The relation between the **volume** of the gas and the **temperature** is established by **Charles's** law. With a **decrease** in the temperature, the volume decreases by **45.7 mL**. Thus, option **c **is correct.

<h3>What is Charle's law?</h3>

**Charle's law **states the direct relation present between the **temperature** and the **volume** of the gas. The law is given as:

V₁ ÷ T₁ = V₂ ÷ T₂

Given,

V₁ = 50 mL

T₁ = 303.15 K

T₂ = 277.15 K

**Substituting** the value the **final** **volume** is calculated as:

50 ÷ 303.15 = V₂ ÷ 277.15

V₂ = (50 × 277.15) ÷ 303.15

**= 45.71 mL**

Therefore, option c. **45.7** **mL **is the final volume.

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Conduction should be the correct answer.

The **molality **of the **solution **is obtained as **0.63 m.**

<h3>What is the freezing point?</h3>

The** freezing point** is the **temperature **at which the liquid is converted into solid.

We know that;

ΔT = 3.5° C

K = 1.86° C/m

i = 3

m = ?

Thus;

ΔT = K m i

m = ΔT/K i

m = 3.5° C/ 1.86° C/m * 3

m = **0.63 m**

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**Answer:**

**The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.**

**Explanation:**

Moles of copper =

According to reaction, 1 mol of copper gives 2 moles of nitrogen dioxide gas.

Then 0.03613 moles of copper will give:

of nitrogen dioxide gas

Moles of nitrogen dioxide gas = n = 0.06326 mol

Pressure of the gas = P

P = Total pressure - vapor pressure of water

P = 726 mmHg - 23.8 mmHg = 702.2 mmHg

P = 0.924 atm (1 atm = 760 mmHg)

Temperature of the gas = T = 25.0°C =298.15 K

Volume of the gas = V

V = 1.68 L

**The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.**