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Answer : The initial concentration of HI and concentration of at equilibrium is, 0.27 M and 0.386 M respectively.
Solution : Given,
Initial concentration of and
= 0.11 M
Concentration of and
at equilibrium = 0.052 M
Let the initial concentration of HI be, C
The given equilibrium reaction is,
Initially 0.11 0.11 C
At equilibrium (0.11-x) (0.11-x) (C+2x)
As we are given that:
Concentration of and
at equilibrium = 0.052 M = (0.11-x)
0.11 - x = 0.052
x = 0.11 - 0.052
x = 0.058 M
The expression of will be,
By solving the terms, we get:
C = 0.27 M
Thus, initial concentration of HI = C = 0.27 M
Thus, the concentration of at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M
Answer:
Explanation:
<u><em>1. First determine the empirical formula.</em></u>
a) Base: 100 g of compound
mass atomic mass number of moles
g g/mol mol
C 26.06 12.011 26.06/12.011 = 2.17
H 13.13 1.008 13.13/1.008 = 13.03
N 60.81 14.007 60.81/14.007 = 4.34
b) Divide every number of moles by the smallest number: 2.17
mass number of moles proportion
C 2.17/2.17 1
H 13.03/2.17 6
N 4.34/2.17 2
c) Empirical formula
d) Mass of the empirical formula
<u><em>2. Molecular formula</em></u>
Since the mass of one unit of the empirical formula is equal to the molar mass of the compound, the molecular formula is the same as the empirical formula: