Answer:
0.001 M OH-
Explanation:
[OH-] = 10^-pOH, so
pOH + pH = 14 and 14 - pH = pOH
14 - 11 = 3
[OH⁻] = 10⁻³ ; [OH-] = 0.001 M OH-
When base is added in order to increase the pH of a solution, it results in MOH salt <u>precipitating first.</u>
Equation :
MOH (s) ⇄ M⁺ (aq) ₊ OH⁻(aq)
K
= [M⁺] [OH⁻]
K
= 1.39×10⁻¹² and [M⁺] = 0.001M
∴ 1.39×10⁻¹² = (0.001M) [OH⁻]
[OH⁻] = 1.004×10⁻¹⁹M
pOH = -log [OH⁻] = 11.9
pH = 14 - pOH
= 14 - 11.9
<u>pH = 2.1</u>
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Answer:
Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).
Explanation:
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In this case, according to the given redox reaction, we rewrite it as a convenient first step:

Next, we assign the oxidation numbers as follows:

Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).
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We have to know final temperature of the gas after it has done 2.40 X 10³ Joule of work.
The final temperature is: 75.11 °C.
The work done at constant pressure, W=nR(T₂-T₁)
n= number of moles of gases=6 (Given), R=Molar gas constant, T₂= Final temperature in Kelvin, T₁= Initial temperature in Kelvin =27°C or 300 K (Given).
W=2.4 × 10³ Joule (Given)
From the expression,
(T₂-T₁)=
(T₂-T₁)= 
(T₂-T₁)= 48.11
T₂=300+48.11=348.11 K= 75.11 °C
Final temperature is 75.11 °C.