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
The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm
<h3>
What is Mole ?</h3>
A mole is a very important unit of measurement that chemists use.
A mole of something means you have 6.023 x 10 ²³ of that thing.
- For 2.15 mol of hydrogen sulphide (H₂S) :
1 mole hydrogen sulphide (H₂S) = 34.08088 grams
Therefore,
2.15 mol of hydrogen sulphide (H₂S) = 34.08088 grams x 2.15 mol
= 73.272 gm
- For 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) ;
1 mol of lead(II) iodide, (PbI₂) = 461.00894 grams
Therefore,
3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) = 461.00894 grams x 3.95 × 10⁻³ mol
= 1.82 gm
Hence,The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm
Learn more about mole here ;
brainly.com/question/21323029
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Incorrect, temperature is directly proportional to the avg. KE of a gas.
Explanation:
here is the answer bae. Feel free to ask for more
the answer is 133
because thats how the water is
