Answer:
40% of registered voters voted
Explanation:
3200 out of 8000,
or :
3200 / 8000
= 0.4
0.4 x 100 = 40
40% of registered voters voted
<u>Answer:</u> The correct answer is Option b.
<u>Explanation:</u>
Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.

For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.
For the given options:
- <u>Option a:</u>

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.
- <u>Option b:</u>

This metal can easily get oxidized to
ion and the standard oxidation potential for this is 0.13 V

- <u>Option c:</u>

This metal can easily get oxidized to
ion and the standard oxidation potential for this is 0.0 V

- <u>Option d:</u>

This metal can easily get oxidized to
ion and the standard oxidation potential for this is -0.80 V

- <u>Option e:</u>

This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.
By looking at the standard oxidation potential of the substances, the substance having highest positive
potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.
From the above values, the correct answer is Option b.
Answer:
1 molecule of oxygen has 2 atoms.
So 2,25 molecules of oxygen have 2 x 2,25 atoms.
2 x 2,25 = 4,5 atoms
There are 4,5 atoms in 2,25 molecules oxygen.
To get the value of ΔG we need to get first the value of ΔG°:
when ΔG° = - R*T*㏑K
when R is constant in KJ = 0.00831 KJ
T is the temperature in Kelvin = 25+273 = 298 K
and K is the equilibrium constant = 4.5 x 10^-4
so by substitution:
∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4
= -19 KJ
then, we can now get the value of ΔG when:
ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]
when ΔG° = -19 KJ
and R is constant in KJ = 0.00831
and T is the temperature in Kelvin = 298 K
and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m
so, by substitution:
ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )
= -40