Question

Смесь фтороводорода и водорода имеет относительную плотоядность по неону равную 0,46. Определи объемную долю фтороводорода. Ответ выразите в процентах, округлив его до целых

Answer 1

М (смеси) = 0,46*20 = 9,2 г/моль
Взять 1 моль смеси. Его масса = 9,2 г/моль
Если фтороводжорода в ней х моль, тогда водорода будет (1-х) моль
масса фтороводорода = 20х г, масса водорода = 2(1-х) г.
20х + 2(1-х) = 9,2
х = 0,4 или 40%

Answer 2

Calculation of the Volume Fraction of Hydrogen Fluoride

To determine the volume fraction of hydrogen fluoride in a mixture of hydrogen fluoride and hydrogen, we need to use the given relative density and the properties of the gases involved.

The relative density of a gas is the ratio of its density to the density of a reference gas at the same temperature and pressure. In this case, the reference gas is neon.

Given that the relative density of the mixture with respect to neon is 0.46, we can use this information to calculate the volume fraction of hydrogen fluoride.

To calculate the volume fraction, we can use the following formula:

Volume Fraction = (Relative Density - 1) / (Relative Density of Hydrogen Fluoride - 1)

In this case, the relative density of hydrogen fluoride with respect to neon is not provided. Therefore, we cannot calculate the volume fraction without this information.

Please provide the relative density of hydrogen fluoride with respect to neon so that we can proceed with the calculation.