Answer:a) 0.1 mole. b) 4g. c) 2% d) 196 mL
Explanation: in 200mL , 0.1mole
mw NaOH = 40g/mol —> 4g in 0.1 mole
4g in 200mL so 2g in 100mL
density NaOH = 1g/mL so if 4g in 200 mL, 4mL , 196 mL water
Answer:
1.5057×10^22 atom
Explanation:
As we
1 mole of argon = 40 g of argon
i.e 40 g of argon = 1 mole of argon
1 g of argon = 1/40 mole of argon
1 mole of argon = 6.023×10^23 atom of argon
1/40 mole if argon = 1/40 ×6.023×10^23
= 1.5057×10^22
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!