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Archy [21]
2 years ago
14

In the reaction Pb + 2Ag+ – Pb2+ + 2Ag, the Ag+ is.

Chemistry
1 answer:
Travka [436]2 years ago
5 0
<h2><u>Question :-</u></h2>

In the reaction Pb + 2Ag+ – Pb2+ + 2Ag, the Ag+ is.

a) Reduced, and the oxidation number changes from +1 to 0.

b) Reduces and the oxidation number changes from +2 to 0.

c) Oxidized and the oxidation number changes from 0 to +1.

d) Oxidized, and the oxidation number changes from +1 to 0.

<h3>____________________</h3>

<h2><u>Solution :-</u></h2>

<h3><u>Given Information :-</u></h3>

  • <u>Reaction ➢</u> Pb + 2Ag⁺ ⟶ Pb²⁺ + 2Ag

<h3><u>To Find :-</u></h3>

  • Whether Ag⁺ is:-

  • a) Reduced, and the oxidation number changes from +1 to 0.

  • b) Reduces and the oxidation number changes from +2 to 0.

  • c) Oxidized and the oxidation number changes from 0 to +1.

  • d) Oxidized, and the oxidation number changes from +1 to 0.

<h3><u>Answer :-</u></h3>

Firstly let us know the meaning of Oxidation & Reduction and Oxidizing & Reducing Agent :-

  1. Oxidation is process of loss of electron(s). Either atom or ion undergoing oxidation, is known as reducing agent.
  2. Reduction is process of gain of electron(s). Either atom or ion undergoing reduction, is known as oxidizing agent.

In this reaction, Ag⁺ gains an electron to get <u>reduced</u> to Ag. The oxidation number changes from +1 to 0. Therefore, Ag+ is an oxidizing agent.

Since, this possibility is given in option a). Reduced, and the oxidation number changes from +1 to 0. Hence, option a is the correct answer.

<h3>____________________</h3>

<h2><u>Final Answer :-</u></h2>

  • Option a. Reduced, and the oxidation number changes from +1 to 0 is correct answer.

<h3>____________________</h3>
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4 years ago
5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g) In a titration experiment, H2O2(aq) reacts with aqueous Mn
Tomtit [17]

Answer:

Oxygen in hydrogen peroxide oxidizes from -1 to 0.

Explanation:

Oxidation is the loss of electrons. The specie which is oxidized has has elevation in its oxidation state as compared in the reactant and the products.

The given reaction is shown below as:

5 H_2O_2_{(aq)} + 2 MnO_4^-_{(aq)} + 6 H^+_{(aq)}\rightarrow 2 Mn^{2+}_{(aq)} + 8 H_2O_{(l)} + 5 O_2_{(g)}

Manganese in MnO_4^- has oxidation state of +7

Manganese in Mn^{2+} has an oxidation state of +2

It reduces from +7 to +2

Oxygen in hydrogen peroxide has an oxidation state of -1.

Oxygen in molecular oxygen has an oxidation of 0.

Thus, oxygen in hydrogen peroxide oxidizes from -1 to 0.

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Answer:

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Name four products of incomplete combustion<br><br> 1 - ____<br> 2 -____<br> 3 -____<br> 4 -_____
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3 years ago
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Ilia_Sergeevich [38]

27) Partial pressure of oxygen: 57.8 kPa

29) Final volume: 80 mL

30) Final volume: 8987 L

31) Due to property of water of being polar, ice floats on water

Explanation:

27)

In a mixture of gases, the total pressure of the mixture is the sum of the partial pressures:

p_T = p_1 + p_2 + ... + p_N

In this problem, the mixture contains 3 gases (helium, carbon dioxide and oxygen). We know that the total pressure is

p_T=201.4 kPa

We also know the partial pressures of helium and carbon dioxide:

P_{He}=125.4 kPa\\P_{CO_2}=18.2 kPa

The total pressure can be written as

p_T=p_{He}+p_{CO_2}+p_{O_2}

where p_{O_2} is the partial pressure of oxygen. Therefore, we find

p_{O_2}=p_T-p_{He}-p_{CO_2}=201.4-125.4-18.2=57.8 kPa

29)

Assuming that the pressure of the gas is constant, we can apply Charle's law, which states that:

"For an ideal gas at constant pressure, the volume of the gas is proportional to its absolute temperature"

Mathematically,

\frac{V}{T}=const.

where

V is the volume of the gas

T is the Kelvin temperature

We can re-write it as

\frac{V_1}{T_1}=\frac{V_2}{T_2}

Here we have:

V_1 = 42 mL (initial volume)

T_1=-89^{\circ}C+273=184 K is the initial temperature

T_2=77^{\circ}C+273=350 K is the final temperature

Solving for V2, we find the final volume:

V_2=\frac{V_1 T_2}{T_1}=\frac{(42)(350)}{184}=80 mL

30)

For this problem, we can use the equation of state for ideal gases, which can be written as

\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}

where in this problem:

p_1 = 102.3 kPa is the initial pressure

V_1=1975 L is the initial volume

T_1=25^{\circ}C+273=298 K is the initial temperature

p_2=21.5 kPa is the final pressure

T_2=12^{\circ}C+273=285 K is the final temperature

And solving for V2, we find the final volume of the balloon:

V_2=\frac{p_1 V_1 T_2}{p_2 T_1}=\frac{(102.3)(1975)(285)}{(21.5)(298)}=8987 L

31)

A molecule of water consists of two atoms hydrogen bond with an atom of oxygen (H_2 O) in a covalent bond.

While the molecul of water is overall neutral, due to the higher electronegativity of the oxygen atom, electrons are slightly shifted towards the oxygen atom; as a result, there is a slightly positive charge on the hydrogen side, and a slightly negative charge on the oxygen side (so, the molecules is said to be polar).

As a consequence, molecules of water attract each other, forming the so-called "hydrogen bonds".

One direct consequence of the polarity of water is that ice floats on liquid water.

Normally, for every substance on Earth, the solid state is more dense than the liquid state. However, this is not true for water, because ice is less dense than liquid water.

This is due to the polarity of water. In fact, when the temperature of water is decreased to freezing point and water becomes ice, the hydrogen bondings "force" the molecules to arrange in a lattice structure, so that the molecules become more spaced when they turn into solid state. As a result, ice occupies more volume than water, and therefore it is less dense, being able to float on water.

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

brainly.com/question/3658563

#LearnwithBrainly

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3 years ago
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