[CO] = 1 mol / 2L = 0.5 M
[
According to the equation:
and by using the ICE table:
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
initial 0.5 0.5 0 0
change -X -X +X +X
Equ (0.5-X) (0.5-X) X X
when Kc = X^2 * (0.5-X)^2
by substitution:
1.845 = X^2 * (0.5-X)^2 by solving for X
∴X = 0.26
∴ [CO2] = X = 0.26
CH4 is an emprirical formula as it shows the simplest ratio of the numbers of different atoms present in the molecule. The empirical formula for CH4 is also the same as the molecular formula.
The other compunds can be simplified so they are not the empirical formula of compounds.
Hope this helps :).
Answer:
2AlCl3 + 3H2SO4 → Al2(SO4)3 + 6HCl
Explanation:
molar mass of methane CH4
= C + 4 H
= 12.0 + 4 x 1.008
= 12.0 + 4.032
= 16.042g/mol
7.31 x 10^25 molecules x <u> 1 mole CH4 </u> = 121.43 moles
6.02 x 10^23 CH4 molecules
121.43 moles CH4 are present.
I just know the ph is between 7 and 8