<u>Answer:</u>
<em>A. 10.25</em>
<em></em>
<u>Explanation:</u>
Pkb =4.77
So pka = 14 - pka = 9.23
Initial 0.50M 0 0
Change -x +x +x
Equilibrium 0.50M-x +x +x
(-x is neglected) so we get
![pH=-log[H^3 O^+]\\\\pH=-log[1.72\times10^{-5}]\\\\pH=4.76](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E3%20O%5E%2B%5D%5C%5C%5C%5CpH%3D-log%5B1.72%5Ctimes10%5E%7B-5%7D%5D%5C%5C%5C%5CpH%3D4.76)
pOH = 14 - pH
= 14 - 4.76
pOH = 9.24 is the answer
Option A - 10.25 is the answer which is close to 9.24
M (NaCl) = n × M = 3.75 × 58.5 =
219.375 g
M (NaCl) = 23 + 35.5 = 58.5 g/mol
<span> Au</span>₂(SeO₄)₃
O = -2 × 4 = -8
Se = + 6
So,
(+6 - 8) = -2
Means (SeO₄) contains -2 charge, Now multiply -2 by 3
-2 ₓ 3 = -6
Means,
Au₂ + (-6) = 0
Au₂ = +6
Or,
Au = 6 / 2
Au = +3
Result:
Au = +3
Se = +6
O = -2
Ni(CN)₂
Cyanide (CN⁻) contains -1 charge,
So,
N = -3
C = +2
Then,
Ni + (-1)₂ = 0
Ni - 2 = 0
Or,
Ni = +2
Result:
N = -3
C = +2
Ni = +2
<span>As the pressure is increased the solubility of the sugar and carbon dioxide is increased. The pressure of combination leaves little to no separation. There would be no discernable difference between the ingredients used to make the saturated solution.</span>
Answer:
I don't know the answer of this question.
