Answer:
°C = °K - 273.
Explanation:
the 1st and 2d rows belong to F°; the last one is wrong; correct answer is °C=°K-273 only.
For example, 0°C=273°K; 10°C=283°K; 100°C=373°K.
According to its formula FeSO4.7H2O
we can get the percent % by the mass of H2O from this formula
%mass of H2O = (mass of water H2O/ mass of the hydrate)x100
when the mass of water = molar mass x 7 = 18 x 7 = 126
and the mass of hydrate (feSO4) = molar mass = 278
So by substitution:
%mass of H2O = (126/278) x 100 = 45%
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!
Molarity = Moles / Liters! So in this instance, if you have 0.45 moles of NaCl, and we convert 300.0 mL to 0.3 L, we would plug in M = 0.45/0.3. The answer would be 1.5 M NaCl!