Answer: 8.2
Explanation:
pH of a solution is - Log [ H+].
pH = - Log [ 6.2 x 10-9 M]
= 9 - 0.7924
pH = 8.24 approx 8.2
We know that density = mass /volume
so
mass=volume*density
volume=mass/density
so c is wrong
Answer:
Reduction
Explanation:
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Mn⁺⁷ +3e⁻ → Mn⁴⁺
Mn gets three electrons , its oxidation state reduced from +7 to +4 so Mn gets reduced.
Examples:
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
PH = -log [H3O+]
4.15 = -log [H3O+]
[H3O+] = 10^(-4.15)
[H3O+]= 7.08 × 10^-5
Answer:- Third choice is correct, 17.6 moles
Solution:- The given balanced equation is:
Al_2(SO_4)_3+6KOH\rightarrow 2Al(OH)_3+3K_2SO_4
We are asked to calculate the moles of potassium hydroxide needed to completely react with 2.94 moles of aluminium sulfate.
From the balanced equation, there is 1:6 mol ratio between aluminium sulfate and potassium hydroxide.
It is a simple mole to mole conversion problem. We solve it using dimensional set up as:
2.94molAl_2(SO_4)_3(\frac{6molKOH}{1molAl_2(SO_4)_3})
= 17.6 mol KOH
So, Third choice is correct, 17.6 moles of potassium hydroxide are required to react with 2.94 moles of aluminium sulfate.
