<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.0275 M
<u>Explanation:</u>
Molarity is calculated by using the equation:
Moles of HI = 0.550 moles
Volume of container = 2.00 L
For the given chemical equation:
<u>Initial:</u> 0.275
<u>At eqllm:</u> 0.275-2x x x
The expression of for above equation follows:
We are given:
Putting values in above expression, we get:
Neglecting the negative value of 'x' because concentration cannot be negative
So, equilibrium concentration of hydrogen gas = x = 0.0275 M
Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M
i know don't know the answerrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr rrrrrrrrrrrrrrrrrrrrr but just guessing is d. sodium
Answer:
Explanation:
<u>1. Molecular chemical equation:</u>
- 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)
<u>2. Mole ratios:</u>
- 2 mol KClO₃ : 2 mol KCl : 3 mol O₂
<u>3. Number of moles of KClO₃</u>
- Number of moles = mass in grams / molar mass
- Molar mass of KClO₃ = 122.55 g/mol
- Number of moles of KClO₃ = 54.3 g / 122.5 g/mol ≈ 0.44308 mol
<u>3. Number of moles of O₂</u>
As per the theoretical mole ratio 2 mol of KClO₃ produce 3 mol of O₂, then set up a proportion to determine how many moles of O₂ will be produced from 0.44038 mol of KClO₃.
- 3 mol O₂ / 2 mol KClO₃ = x / 0.44038 mol KClO₃
- x = (3 / 2) × 0.44308 mol O₂ = 0.6646 mol O₂
Round to 3 significant figures: 0.665 mol of O₂ ← answer