How many moles of argon are in 30.6L at 28K and 658 mmHg?
1 answer:
Answer : The number of moles of argon gas is, 11.5 mol
Explanation :
To calculate the moles of argon we are using ideal gas equation as:
where,
P = pressure of argon gas = 658 mmHg = 0.866 atm (1 atm = 760 mmHg)
V = volume of argon gas = 30.6 L
n = number of moles of argon gas = ?
R = gas constant = 0.0821 L.atm/mol.K
T = temperature of argon gas = 28 K
Now put all the given values in the above formula, we get:
Therefore, the number of moles of argon gas is, 11.5 mol
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Answer:
6,45mmol/L of NaOH you need to add to reach this pH.
Explanation:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ <em>pka = 4,74</em>
Henderson-Hasselbalch equation for acetate buffer is:
5,0 = 4,74 + log₁₀
Solving:
1,82 = <em>(1)</em>
As total concentration of acetate buffer is:
10 mM = [CH₃COOH] + [CH₃COO⁻] <em>(2)</em>
Replacing (2) in (1)
<em>[CH₃COOH] = 3,55 mM</em>
And
<em>[CH₃COO⁻] = 6,45 mM</em>
Knowing that:
<em>CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O</em>
Having in the first 10mmol/L of CH₃COOH, you need to add <em>6,45 mmol/L of NaOH. </em>to obtain in the last 6,45mmol/L of CH₃COO⁻ and 3,55mmol/L of CH₃<em>COOH </em>.
I hope it helps!