Mass of Co(NO₃)₂ = 1.95 g
V KOH = 0.350 L
[KOH] = 0.220 M
Kf = 5.0 x 10⁹
molar mass of Co(NO₃)₂ = 182.943 g/mol
so [Co(NO₃)₂] = 1.95 / (0.350 * 182.943) = 0.03045 M
[Co²⁺] = 0.03045 M
[OH⁻] = 0.22 M
chemical reaction:
Co²⁺(aq) + 4 OH⁻ ⇄ Co(OH)₄²⁻
I (M) 0.03045 0.22 0
C (M) - 0.03045 - 4 (0.03045) 0.03045
E (M) - x 0.22 - 4(0.03045) 0.03045
= 0.0982
Kf = [Co(OH)₄²⁻] / [Co⁺²][OH⁻]⁴
5.0 x 10⁹ = (0.03045) / x (0.0982)⁴
x = 6.5489 x 10⁻⁸
at equilibrium:
[Co²⁺] = 6.54 x 10⁻⁸
[OH⁻] = 0.0982 M
[Co(OH)₄²⁻] = 0.03045 M
Answer:
M
Explanation:
Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.
Partial pressure = Henry's constant x molar concentration
0.21 = Hx1.38x
H = 
H = 152.17 atm/M
For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:
P = 0.21*0.875 = 0.1837 atm
Then, the molar concentration [O2], will be:
P = Hx[O2]
0.1837 = 152.17x[O2]
[O2] = 0.1837/15.17
[O2] =
M
D milk turning sour
the other options are physical changes