Answer:
24x10³
Explanation:
2CO₂(g) + 4H₂O(g) → 2CH₃OH(l) + 3O₂ (g)
The equilibrium constant for this reaction is:
Kc =
The expression of [CH₃OH] is left out as it is a pure liquid.
Now we <u>convert the given masses of the relevant species into moles</u>, using their <em>respective molar masses</em>:
- CO₂ ⇒ 3.28 g ÷ 44 g/mol = 0.0745 mol CO₂
- H₂O ⇒ 3.86 g ÷ 18 g/mol = 0.214 mol H₂O
- O₂ ⇒ 2.80 g ÷ 32 g/mol = 0.0875 mol O₂
Then we calculate the concentrations:
- [CO₂] = 0.0745 mol / 7.5 L = 0.0099 M
- [H₂O] = 0.214 mol / 7.5 L = 0.0285 M
- [O₂] = 0.0875 mol / 7.5 L = 0.0117 M
Finally we <u>calculate Kc</u>:
- Kc = = 24x10³
Answer:
what does that supposed to mean?
Answer:
23
Explanation:
Simply remove the brackets and add 14 and 9
The question is missing the number of moles of carbon dioxide that will react with water.
I will work the problem with an arbitrary amount of carbon dioxide to show how to solve it.
For this, I will take 2.40 moles of carbon dioxide.
Answer:
- <u>The number of moles of oxygen atoms produced is equal to the number of moles of carbon dioixide that react: 2.40 moles of oxygen.</u>
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- 2.40 has 3 significant figures.
Explanation:
<u>1) Word equation:</u>
- carbon dioxide gas + liquid water (in the presence of light) → aqueous glucose + oxygen gas
<u>2) Balanced chemical equation</u>
- 6CO₂ (g) + 6H₂O (l) → C₆H₁₂O₆ (aq) + 6O₂(g)
<u>3) Mole ratios</u>
- 6 mol CO₂ (g) : 6 mol H₂O (l) : 1 mol C₆H₁₂O₆ (aq) : 6 mol O₂(g)
<u>4) Set a proportion:</u>
It is assumed that there is plenty liquid water (excess reactant), so you can set a proportion with the number of moles of carbon dioxide:
- 6 mol CO₂ / 6 mol O₂ = 2.40 mol CO₂ / x
From which, x = 2.40 mol O₂
So, the number of moles of oxygen produced is equal to the number of moles of carbon dioxide that react.
Since the number of moles of reactant has 3 significant figures, and the stoichiometric coefficients are considered exact, the answer also has 3 significant figures.
Answer:
I think c
Explanation:
<h2>I HOPE IT'S HELP</h2>