Question

Газовая смесь, состоящая из летучих водородных соединений азота и углерода, занимает объем 44,8дм3 и имеет массу 32.5г. Рассчитайте массы соединений в смеси. Пожалуйста срочно​

Answer 1

Ответ:

$\frac{44.8}{22.4} = 2 \: мол \\ \\ N2 = 28x \: \: \: \: \: \: \: \: \: N2 = 2y \\ \\ 28x + 2y = 32.5 \\ x + y = 2 \\ \\ x = 30.8 \\ y = 1.8 \\ \\ N2 = 30.82 \\ N2 = 1.8г$

Answer 2

Ответ:

44,8/22,4=2мол

N2=28x H2=2y

{ 28x+2y=32,5 x=30,8 y=1,8

{ x+y=2

N2=30,8г

`H2=1,8г

Answer 3

I will answer in detail. The gas mixture, consisting of volatile hydrogen compounds of nitrogen and carbon, occupies a volume of 44.8 dm3 and has a mass of 32.5 g. Calculate the masses of the compounds in the mixture. Please urgently.

To solve this problem, we need to assume that the gas mixture consists of ammonia (NH3) and methane (CH4), which are the simplest volatile hydrogen compounds of nitrogen and carbon. We also need to use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the amount of substance, R is the universal gas constant, and T is the temperature. We can assume that the gas mixture is at standard conditions, which means that P = 101.3 kPa and T = 273.15 K. R can be taken as 8.314 J/(mol*K).

Let x be the amount of ammonia and y be the amount of methane in the gas mixture. Then we can write the following system of equations:

17x + 16y = 32.5 (1) - this is the mass balance equation, based on the molar masses of ammonia (17 g/mol) and methane (16 g/mol).

x + y = 44.8 / 22.4 (2) - this is the volume balance equation, based on the fact that one mole of any ideal gas occupies 22.4 dm3 at standard conditions.

Solving this system of equations, we get x = 0.5 mol and y = 1.5 mol. Therefore, the mass of ammonia in the gas mixture is 0.5 * 17 = 8.5 g, and the mass of methane is 1.5 * 16 = 24 g.