Complete question:
Consider the reaction.
At equilibrium at 600 K, the concentrations are as follows.
2HF -----> H₂ + F₂
[HF] = 5.82 x 10-2 M
[H2] = 8.4 x 10-3 M
[F2] = 8.4 x 10-3 M
What is the value of Keq for the reaction expressed in scientific notation?
2.1 x 10-2
2.1 x 102
1.2 x 103
1.2 x 10-3
Answer:
2.1 × 10^-2
Explanation:
Kequilibrum(Keq) = product/reactant
Equation for the reaction :
2HF -----> H₂ + F₂
Therefore,
Keq = [H2][F2] / [HF]^2
Keq = [8.4 x 10-3][8.4 x 10-3] / [5.82 x 10-2]^2
Keq = [70.56 × 10^(-3 + - 3)]/[33.8724 × 10^(-2×2)]
Keq = [70.56 × 10^-6] / [33.8724 × 10^-4]
Keq = 2.0665 × 10^(-6 - (-4))
Keq = 2.0665 × 10^(-6 + 4)
Keq = 2.1 × 10^-2
Explanation:
require an emergency support immediately
<span>Answer:
W must be 5-chloro-2-methylpentane. It can give only 4-methy-1-pentene (Y) upon dehydrohalogenation:
X must be 4-chloro-2-methylpentane. Dehydrohalogenation yields both Y and 4-methyl-2-pentene. (Z)</span>
Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
To get the value of ΔG we need to get first the value of ΔG°:
when ΔG° = - R*T*㏑K
when R is constant in KJ = 0.00831 KJ
T is the temperature in Kelvin = 25+273 = 298 K
and K is the equilibrium constant = 4.5 x 10^-4
so by substitution:
∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4
= -19 KJ
then, we can now get the value of ΔG when:
ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]
when ΔG° = -19 KJ
and R is constant in KJ = 0.00831
and T is the temperature in Kelvin = 298 K
and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m
so, by substitution:
ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )
= -40