Molar mass of CO2 = 44.01 g/mol
Answer: option C) II < III < I
i.e [OH−] < [H3O+] < I
Explanation:
First, obtain the pH value of I and II, then compare both with III.
For I
Recall that pH = -log (H+)
So pH3O = -log (H3O+)
= - log (1x10−5)
= 4
For II
pOH = - log(OH-)
= - log(1x10−10)
= 9
For III
pH = 6
Since, pH range from 1 to 14, with values below 7 to be acidic, 7 to be neutral, above 7 to be alkaline: then, 9 < 6 < 4
Thus, the following solutions from least acidic to most acidic is II < III < I
Solution with a pH of 3 has 10⁻³ moles of H⁺, solution with a pH of 4 has 10⁻⁴ moles of H⁺ and solution with a pH of 5 has 10⁻⁵ moles of H⁺ (in dm³) so the solution with a pH of 3 has 10 times more H⁺ ions than the solution with a pH of 4 and 100 times more H⁺ ions than the solution with a pH of 5.
Answer:
The element that has been oxidized is the N
Explanation:
Zn²⁺(aq) + NH₄⁺(aq) → Zn(s) + NO₃⁻(aq)
See all the oxidation states:
Zn²⁺ → acts with +2
In ammonia, H acts with +1 and N with -3
Zn(s), acts with 0. In all the elements in ground state, the oxidation state is 0.
Zn changed from 2+ to 0. The oxidation number, has decreased.
This element has been reduced.
NO₃⁻ (aq) it's a ion, from nitric acid.
N acts with +5
O acts with -2
The global charge is -1
The N, has increased the oxidation state, so this element is the one oxidized.
