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Georgia [21]
3 years ago
9

g Select the correct statements. I. Reduction is the loss of electrons, and oxidation is the gain of electrons II. Reduction is

the gain of electrons, and oxidation is the loss of electrons III. The reactant that is reduced is also called the reducing agent, and the reactant that is oxidized is called the oxidizing agent. IV. The reactant that is oxidized is also called the reducing agent, and the reactant that is reduced is called the oxidizing agent. V. The sum of all the oxidation numbers in a compound is equal to the charge of that compound. A. I and III only B. II and IV only C. I, III, and V only D. II, IV, and V only E. V only
Chemistry
1 answer:
Karolina [17]3 years ago
6 0

<u>Answer:</u> The correct answer is Option D.

<u>Explanation:</u>

Reduction reaction is defined as the reaction in which a substance gains electrons. Here, the oxidation state of the substance decreases.

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

Oxidizing agents are the agents that helps in the oxidation of other substance and itself gets reduced. These agents undergoes reduction reactions.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. Here, oxidation state of the substance increases.

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

Reducing agents are the agents that helps in the reduction of the other substance and itself gets oxidized. These agents undergoes reduction reactions.

Oxidation state is the number which is given to an atom when it looses or gains electron. It is written as a superscript. In a compound, the total charge is equal to the sum of the charges of all atoms in that compound. <u>For Example:</u> In MnO_4^-, manganese has +7 oxidation number and oxygen has -2.

So, the charge on the compound = [=7+(4\times (-2))]=-1

Hence, the correct answer is Option D.

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Using henry's law, calculate the molar concentration of o2 in the surface water of a mountain lake saturated with air at 20 ∘c a
s2008m [1.1K]

Answer: The molar concentration of oxygen gas in water is 1.43\times 10^{-7} mol/L.

Explanation:

Partial pressure of the O_2gas = 685 torr = 0.8905 bar

1 torr = 0.0013 bar

According Henry's law:

p_{o_2}=K_H\times\chi_{O_2}

Value of Henry's constant of oxygen gas at 20 °C in water = 34860 bar

0.0013=34860 bar\times \chi_{O_2}

\chi_{O_2}=\frac{0.8905 bar}{34680 bar}=2.56\times 10^{-5}

Let the number of moles of O_2 gas in 1 liter water be n.

1 Liter water = 1000 g of water

Moles of water in 1 L n_w=\frac{1000 g}{18 g/mol}=55.55 mol

\chi_{O_2}=\frac{n}{n+n_w}

2.56\times 10^{-5}=\frac{n}{n+55.55}

n=1.43\times 10^{-7} moles

Molarity=\frac{\text{Moles of}O_2}{Volume}

Molar concentration of oxygen gas in 1 L of water:

=\frac{1.43\times 10^{-7} moles}{1 L}=1.43\times 10^{-7} mol/L

The molar concentration of oxygen gas in water is 1.43\times 10^{-7} mol/L.

4 0
3 years ago
Water molecules are polar because the?
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Answer:

Water (H2O) is polar due to the  bent shape of the molecule. The shape means most of the negative charge from the oxygen is one one side of the molecule and the positive charge of the hydrogen atoms is on the other side of the molecule. This is an example of polar covalent chemical bonding.

Hope this helps!

6 0
3 years ago
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Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

C(s)+H_2O(l)\leftrightharpoons CO(g)+H_2(g)

A. C is added to the reaction mixture.

If the concentration of reactant specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of reactant specie occurs. So, on increasing C ,equilibrium will shift in right direction.

B. H_2O is condensed and removed from the reaction mixture.

If the concentration of reactant specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of reactant specie occurs. So, on removing water vapor ,the equilibrium will shift in left direction.

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If the concentration of product specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of product specie occurs. So, on increasing CO, the equilibrium will shift in left direction.

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5 0
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If the Dry bulb reads 25 degrees Celsius and the wet bulb reads 22
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Answer:

Relative humidity is low .

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