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Galina-37 [17]
3 years ago
12

Look at the image please. I need the help as soon as possible.

Chemistry
1 answer:
boyakko [2]3 years ago
5 0

Answer:

It's the first option

Explanation:

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Under standard-state conditions, which of the following species is the best reducing agent? a. Ag+ b. Pb c. H2 d. Ag e. Mg2+
eimsori [14]

<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.

X\rightarrow X^{n+}+ne^-

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>  Ag^+

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.

  • <u>Option b:</u>  Pb

This metal can easily get oxidized to Pb^{2+} ion and the standard oxidation potential for this is 0.13 V

Pb\rightarrow Pb^{2+}+2e^-;E^o_{(Pb/Pb^{2+})}=+0.13V

  • <u>Option c:</u>  H_2

This metal can easily get oxidized to H^{+} ion and the standard oxidation potential for this is 0.0 V

H_2\rightarrow 2H^++2e^-;E^o_{(H_2/H^{+})}=0.0V

  • <u>Option d:</u>  Ag

This metal can easily get oxidized to Ag^{+} ion and the standard oxidation potential for this is -0.80 V

Ag\rightarrow Ag^{+}+e^-;E^o_{(Ag/Ag^{+})}=-0.80V

  • <u>Option e:</u>  Mg^{2+}

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 E^o 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.

8 0
3 years ago
You have 4 moles of oxygen gas in a flask. 4 moles of helium gas is added. What happens to the total pressure of the gases in th
ASHA 777 [7]

Answer: The correct option is (c). The total pressure doubles.

Solution:

Initially,  only 4 moles of oxygen gas were present in the flask.

p_{O_2}=Tp_1\times X_{O_2}  (X_{O_2}=\frac{4}{4}) ( according to Dalton's law of partial pressure)

p_{O_2}=Tp_1\times 1=Tp_1....(1)

Tp_1= Total pressure when only oxygen gas was present.

Final total pressure when 4 moles of helium gas were added:

X'_{O_2}=\frac{4}{8}=\farc{1}{2},X_{He}=\frac{4}{8}=\frac{1}{2}

partial pressure of oxygen in the mixture :

Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.

p_{O_2}=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

Tp_2= Total pressure of the mixture.

from (1)

Tp_1=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

On rearranging, we get:

Tp_2=2\times Tp_1

The new total pressure will be twice of initial total pressure.

7 0
3 years ago
What point A on the phase diagram called?
Mariana [72]

Answer:

D) the critical point

Explanation:

Point A is the critical point in phase diagram. This is the highest temperature and pressure at which a pure material can exist in vapor/liquid equilibrium. Pretty cool!

6 0
3 years ago
Which one of the following polymer is prepared by condensation polymerization?
nikitadnepr [17]

The answer is Darcon my friend, have a fantastic day

8 0
3 years ago
Read 2 more answers
The placement of carbonyl group of a kerose sugar is at second-carbon only (b) fir + carbon only () first and fast carbon (d) li
olasank [31]

Answer:

(d)

Explanation:

Carbonyl group can be the placement of kerosene sugar

6 0
3 years ago
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