Answer:
The correct answer is V1/T1=V2/T2.
Explanation:
Just took the test
Actions form positive ions while anions forms negative
<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) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.


There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6