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2 years ago

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.

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

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

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

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 :-

Oxidation is process of loss of electron(s). Either atom or ion undergoing oxidation, is known as reducing agent.
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.

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

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

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You'll want to add three amounts of heat.

(1) Specific heat of lowering the temperature from -135°C to the melting point -114°C
(2) Latent heat of fusion/melting
(3) Specific heat of elevating the temperature from -114°C to -50°C

(1) E = mCΔT = (25 g)(0.97 J/g·°C)(1 kJ/1000 J)(-114 - -135) = 0.509 kJ
(2) E = mΔH = (25 g)(5.02 kJ/mol)(1 mol/46.07 g ethanol) = 2.724 kJ
(3) E = mCΔT = (25 g)(2.3 J/g·°C)(1 kJ/1000 J)(-50 - -114) = 3.68 kJ

<em>Summing up all energies, the answer is 6.913 kJ.</em>

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3 years ago

CHEM HELP!

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So let's convert this amount of mL to grams:

$\frac{13.6g}{1mL}*1.2mL=16.32g$

Then we need to convert to moles using the molar weight found on the periodic table for mercury (Hg):

$\frac{1mole}{200.59g}*16.32g=8.135*10^{-2}mol$

Then we need to convert moles to atoms using Avogadro's number:

$\frac{6.022*10^{23}atoms}{1mole} *[8.135*10^{-2}mol]=4.90*10^{22}atoms$

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3 years ago

A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an

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Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

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3 years ago

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Answer: 250

Explanation:

You work this problem by using proportions.

A proportion is the equalization of two ratios.

Here you assume that the ratio of fish with tags to total fish that you catch is the same than the ratio of fish with tags to total fish in the pond.

Mathematically:

5 fish with tag / 25 fish = 50 fish with tag / x

5 / 25 = 50 / x

Solve for x:

Multiplication property of equality: x × 5 = 50 × 25
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3 years ago

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