Question

Обчисли об єм (н.у.) повітря, шо витрачається на окиснення ртуті масою 40,2 г.

Answer 1

Ответ:

10,7 л

Объяснение:

в фото

Answer 2

Calculating the Volume of Air Required for the Oxidation of Mercury

To calculate the volume of air required for the oxidation of mercury, we can use the ideal gas law equation, which is expressed as:

PV = nRT

Where: - P = pressure - V = volume - n = number of moles of gas - R = ideal gas constant - T = temperature

First, we need to determine the number of moles of mercury (Hg) involved in the reaction. The molar mass of mercury is 200.59 g/mol.

Given: - Mass of mercury (m) = 40.2 g - Molar mass of mercury (M) = 200.59 g/mol

We can calculate the number of moles (n) using the formula:

n = m / M

n = 40.2 g / 200.59 g/mol n ≈ 0.20 mol

Now, we can use the ideal gas law to calculate the volume of air required. The molar volume of an ideal gas at standard temperature and pressure (STP) is approximately 22.4 L/mol.

V = nRT / P

At STP, the pressure (P) is 1 atm and the temperature (T) is 273 K.

V = (0.20 mol)(0.0821 L·atm/mol·K)(273 K) / 1 atm V ≈ 4.49 L

Therefore, the volume of air required for the oxidation of 40.2 g of mercury is approximately 4.49 liters.