Question

Встанови кількість речовини і який об'єм (л) за нормальних умов займе чадний газ CO, масою 2,8 г.

Answer 1

Ответ:

n(CO) = m / M = 2,8 / 12 + 16 = 2,8г / 28г/моль = 0,1 моль

V(CO) = n * Vm = 0,1моль * 22,4л/моль = 2,24л

V(CO) = 2,24л

Объяснение:

Answer 2

Calculating the Volume of Carbon Monoxide Gas

To calculate the volume of carbon monoxide gas (CO) at standard conditions, we can use the ideal gas law equation:

PV = nRT

Where: - P = pressure (in atmospheres) - V = volume (in liters) - n = number of moles - R = ideal gas constant (0.0821 L·atm/mol·K) - T = temperature (in Kelvin)

First, we need to find the number of moles of CO gas using its mass and molar mass.

Finding the Number of Moles of CO Gas

The molar mass of carbon monoxide (CO) is the sum of the atomic masses of carbon and oxygen: - Atomic mass of C = 12.01 g/mol - Atomic mass of O = 16.00 g/mol

So, the molar mass of CO = 12.01 + 16.00 = 28.01 g/mol.

Given mass of CO = 2.8 g, we can calculate the number of moles using the formula: moles = mass / molar mass

moles = 2.8 g / 28.01 g/mol

moles ≈ 0.1 mol

Calculating the Volume at Standard Conditions

Now that we have the number of moles, we can use the ideal gas law to find the volume at standard conditions. At standard conditions, the pressure is 1 atmosphere and the temperature is 273.15 K.

Plugging in the values: PV = nRT V = (nRT) / P V = (0.1 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm

V ≈ 2.25 L

So, under normal conditions, 2.8 grams of carbon monoxide gas would occupy approximately 2.25 liters.