Answer:
8.08 × 10⁻⁴
Explanation:
Let's consider the following reaction.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
The initial concentration of phosgene is:
M = 2.00 mol / 1.00 L = 2.00 M
We can find the final concentrations using an ICE chart.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
I 2.00 0 0
C -x +x +x
E 2.00 -x x x
The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.
The concentrations at equilibrium are:
[COCl₂] = 2.00 -x = 1.96 M
[CO] = [Cl₂] = 0.0398 M
The equilibrium constant (Keq) is:
Keq = [CO].[Cl₂]/[COCl₂]
Keq = (0.0398)²/1.96
Keq = 8.08 × 10⁻⁴
<span>Answer:
Zn(2+) + 2e(-) -------> Zn
1 mole of Zn is deposited by 2F of electricity ...
so 48.9 mole of Zn will be deposited by 48.9 X 2F = 97.8 F of electricity...
as 1F = 96500 C
so 97.8 F = 97.8 X 96500 = 9437700 C of electricity...</span>
Answer:
1) The Kelvin temperature cannot be negative
2) The Kelvin degree is written as K, not ºK
Explanation:
The temperature of an object can be written using different temperature scales.
The two most important scales are:
- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).
- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.
The expression to convert from Celsius degrees to Kelvin is:
Therefore in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:
1) The Kelvin temperature cannot be negative
2) The Kelvin degree is written as K, not ºK
Molarity after dilution : 0.0058 M
<h3>Further explanation
</h3>
The number of moles before and after dilution is the same
The dilution formula
M₁V₁=M₂V₂
M₁ = Molarity of the solution before dilution
V₁ = volume of the solution before dilution
M₂ = Molarity of the solution after dilution
V₂ = Molarity volume of the solution after dilution
M₁=0.1 M
V₁=6.11
V₂=105.12