Answer:
D is the answer to your question in my opinion
Answer:
<h2>377 kPa</h2>
Explanation:
The original pressure can be found by using the formula for Boyle's law which is
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we're finding the original pressure
150 kPa = 150,000 Pa
We have
We have the final answer as
<h3>377 kPa</h3>
Hope this helps you
Answer:
C co2 2co enthalpy
2 Answers. Ernest Z. The standard enthalpy of formation of carbon monoxide is -99 kJ/mol.
Any salt, even though it reacts with water, will precipitate out completely when the water is completely evaporated. If you start with 50 grams of salt, you will end up with 50 grams of salt.
Answer:
0.098 moles H₂S
Explanation:
The reaction that takes place is
- 2H₂(g) + S₂(g) ⇄ 2H₂S(g) keq = 7.5
We can express the equilibrium constant as:
- keq = [H₂S]² / [S₂] [H₂]² = 7.5
With the volume we can <u>calculate the equilibrium concentration of H₂</u>:
- [H₂] = 0.072 mol / 2.0 L = 0.036 M
<em>The stoichiometric ratio</em> tells us that <u>the concentration of S₂ is half of the concentration of H₂</u>:
- [S₂] = [H₂] / 2 = 0.036 M / 2 = 0.018 M
Now we <u>can calculate [H₂S]</u>:
- 7.5 = [H₂S]² / (0.018*0.036²)
So 0.013 M is the concentration of H₂S <em>at equilibrium</em>.
- This would amount to (0.013 M * 2.0 L) 0.026 moles of H₂S
- The moles of H₂ at equilibrium are equal to the moles of H₂S that reacted.
Initial moles of H₂S - Moles of H₂S that reacted into H₂ = Moles of H₂S at equilibrium
Initial moles of H₂S - 0.072 mol = 0.026 mol
Initial moles of H₂S = 0.098 moles H₂S
