Answer:
M KIO3 = 1.254 mol/L
Explanation:
∴ w KIO3 = 553 g
∴ mm KIO3 = 214.001 g/mol
∴ volumen sln = 2.10 L
⇒ mol KIO3 = (553 g)×(mol/210.001 g) = 2.633 mol
⇒ M KIO3 = (2.633 mol KIO3 / (2.10 L sln)
⇒ M KIO3 = 1.254 mol/L
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!
Barium Chloride
Aluminum Iodide
Lithium Phosphide
Sodium Nitride
Potassium Sulfide
Aluminum Oxide
Sodium Oxide
Rubidium Bromide
Calcium Phosphide
hope this helps for the names
It would be 4.2, hope this helps.
Physical properties are those which can be observed without any change in composition of the substance. Hence, a is the answer.