Answer:
Explanation:
To calculate pH you need to use Henderson-Hasselbalch formula:
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Where HA is the acid concentration and A⁻ is the conjugate base concentration.
The equilibrium of acetic acid is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75
Where <em>CH₃COOH </em>is the acid and <em>CH₃COO⁻ </em>is the conjugate base.
Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:
pH = 4,75 + log₁₀ ![\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOO%5E-%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
a) The pH is:
pH = 4,75 + log₁₀ ![\frac{[2 mol]}{[2 mol]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2%20mol%5D%7D%7B%5B2%20mol%5D%7D)
<em>pH = 4,75</em>
<em></em>
b) The pH is:
pH = 4,75 + log₁₀ ![\frac{[2 mol]}{[1mol]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2%20mol%5D%7D%7B%5B1mol%5D%7D)
<em>pH = 5,05</em>
<em></em>
I hope it helps!
Answer:
2.82 L
T₁ = 303 K
T₂ = 263 K
The final volume is smaller.
Explanation:
Step 1: Given data
- Initial temperature (T₁): 30 °C
- Initial volume (V₁): 3.25 L
- Final temperature (T₂): -10 °C
Step 2: Convert the temperatures to Kelvin
We will use the following expression.
K = °C + 273.15
T₁: K = 30°C + 273.15 = 303 K
T₂: K = -10°C + 273.15 = 263 K
Step 3: Calculate the final volume of the balloon
Assuming constant pressure and ideal behavior, we can calculate the final volume using Charles' law. Since the temperature is smaller, the volume must be smaller as well.
V₁/T₁ = V₂/T₂
V₂ = V₁ × T₂/T₁
V₂ = 3.25 L × 263 K/303 K = 2.82 L
Answer:
B.
Explanation:
One mole is the amount of substance that contain the Avogadro number which is equal to 6.022×10^23 atom, molecules or ions.
Answer:
a. 58.5 g/mol
b. 0.1 mol
Explanation:
a.
The molar mass of Na is 23.0 g/mol. The molar mass of Cl is 35.5 g/mol. The molar mass of NaCl is:
M(Na) + M(Cl) = 23.0 g/mol + 35.5 g/mol = 58.5 g/mol
b. A healthy adult should eat no more than 6 g of salt in one day. The moles corresponding to 6 g of NaCl are:
6 g × (1 mol/58.5 g) = 0.1 mol
