A 6.000L tank at 19.2°C is filled with 18.0g of carbon monoxide gas and 10.6g of chlorine pentafluoride gas. You can assume both
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The H⁺ ion concentration can be calculated from pH values using the following equation:
1.) Given pH = 2
Using the above equation, 2 = - log [H⁺]
Therefore, [H⁺] = 10⁻² mol/L
2.) Given pH = 6
Using the same equation, we have 6 = - log [H⁺]
Hence, [H⁺] = 10⁻⁶ mol/L
3.) Taking the ratio of [H⁺] for pH = 2 and pH = 6, we have
= 10⁴
So, there are 10,000 times more H⁺ ions in a solution of pH = 2 than that of pH = 6.
Answer:
9.47 mL
Explanation:
The reaction that takes place is:
- 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
First we <u>calculate how many KOH moles reacted</u>, using <em>the given concentration and volume of KOH solution</em>:
- 0.061 mol/L = 0.061 mmol/mL
- 0.061 mmol/mL * 26.7 mL = 1.6287 mmol KOH
Then we <u>convert KOH moles into H₂SO₄ moles</u>, using the <em>stoichiometric coefficients</em>:
- 1.6287 mmol KOH *
= 0.8144 mmol H₂SO₄
Finally we <u>calculate the required volume of the H₂SO₄ solution</u>, using<em> the number of moles and given concentration</em>:
- 0.8144 mmol ÷ 0.086 mmol/mL = 9.47 mL