Answer:
12.09 L
Explanation:
Step 1: Convert 826.1 mmHg to atm
We will use the conversion factor 760 mmHg = 1 atm.
826.1 mmHg × 1 atm/760 mmHg = 1.087 atm
Step 2: Convert 427.8 J to L.atm
We will use the conversion factor 101.3 J = 1 L.atm.
427.8 J × 1 L.atm/101.3 J = 4.223 L.atm
Step 3: Calculate the change in the volume
Assuming the work done (w) is 4.223 L.atm against a pressure (P) of 1.087 atm, the change in the volume is:
w = P × ΔV
ΔV = w/P
ΔV = 4.223 L.atm/1.087 atm = 3.885 L
Step 4: Calculate the final volume
V₂ = V₁ + ΔV
V₂ = 8.20 L + 3.885 L = 12.09 L
The answer to your question is a 12
Answer:
74
Explanation:
we know,
atomic number = number of protons
so,
atomic number = 74
Data Given:
% w/w = 5 %
Solution weight = 1500 g
Solute weight = ?
Formula Used:
% w/w = (Mass of Solute / Mass of Solution) × 100
Solving for Mass of Solute,
Mass of Solute = (% w/w × Mass of Solution) ÷ 100
Mass of Solute = (5 × 1500 g) ÷ 100
Mass of Solute = 75 g K₂SO₄
Answer:
1, 2, and 3 are true.
Explanation:
The Henderson-Hasselbalch equation is:
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
- If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If you know pH and pka:
10^(pH-pka) = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
The ratio will be: 10^(pH-pka)
- At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
0 = log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
10^0 = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
1 = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
As ratio is 1, [conjugate base] = [acid] in solution.
- At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If pH >> pKa, 10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]
- At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If pH << pKa, 10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]
I hope it helps!