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3 years ago

Given the standard enthalpy changes for the following two reactions:

Chemistry

2 answers:

koban [17]3 years ago

8 0

Answer:

ΔH° = -186.2 kJ

Explanation:

Step 1: Data given

(1) Sn(s) + Cl2(g) ⇆ SnCl2(s) ΔH° = -325.1 kJ

(2) Sn(s) + 2Cl2(g) ⇆ SnCl4(l) ΔH° = -511.3 kJ

Step 2: The balanced equation

SnCl2(s) + Cl2(g) ⇆ SnCl4(l)

Step 3: Calculate the standard enthalpy change for the reaction

We have to take the reverse equation of the first reaction ( Because we need SnCl2 as reactant)

SnCl2(s) ⇆ Sn(s) + Cl2(g) ΔH° = 325.1 kJ

Then we add the second equation to this new one

SnCl2(s) ⇆ Sn(s) + Cl2(g) ΔH° = 325.1 kJ

Sn(s) + 2Cl2(g) ⇆ SnCl4(l) ΔH° = -511.3 kJ

SnCl2(s) + Sn(s) + 2Cl2(g) ⇆ Sn(s) + Cl2(g) + SnCl4(l)

SnCl2(s) + Cl2(g) ⇆ SnCl4(l)

ΔH° = 325.1 kJ + (-511.3kJ)

ΔH° = 325.1 kJ - 511.3 kJ

ΔH° = -186.2 kJ

lawyer [7]3 years ago

7 0

Answer:

ΔH° = -186.2 kJ

Explanation:

Hello,

This case in which the Hess method is applied to compute the required chemical reaction. Thus, we should arrange the given first two reactions as:

(1) it is changed as:

SnCl2(s) --> Sn(s) + Cl2(g)...... ΔH° = 325.1 kJ

That is why the enthalpy of reaction sign is inverted.

(2) remains the same:

Sn(s) + 2Cl2(g) --> SnCl4(l)......ΔH° = -511.3 kJ

Therefore, by adding them, we obtain the requested chemical reaction:

(3) SnCl2(s) + Cl2(g) --> SnCl4(l)

For which the enthalpy change is:

ΔH° = 325.1 kJ - 511.3 kJ

ΔH° = -186.2 kJ

Best regards.

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3 years ago

A solution of NaF is added dropwise to a solution that is 0.0144 M in Ba2+. When the concentration of F- exceeds ________ M, BaF

BARSIC [14]

Answer:

When the concentration of F- exceeds 0.0109 M, BaF2 will precipitate.

Explanation:

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3 years ago

Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring

Lady_Fox [76]

Answer:

The standard cell potential is 1.40 V. The correct option is the option D (+1.40 V)

Explanation:

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

An oxidizing element or oxidizing agent is one that reaches a stable energy state as a result of which the oxidant is reduced and gains electrons. The oxidizing agent causes oxidation of the reducing agent generating the loss of electrons of the substance and, therefore, oxidizes in the process.

In other words, the oxidizing agent is that chemical species that in a redox process accepts electrons released by the reducing agent and, therefore, is reduced in said process. The oxidizing agent is reduced because, upon receiving electrons from the reducing agent, a decrease in the value of the charge or oxidation number of one of the atoms of the oxidizing agent is induced .

Electrochemical cells, galvanic cells or batteries are called devices that are capable of transforming chemical energy originated in a spontaneous redox process into electrical energy.

The cellular potential is generally in standard conditions, that is, 1 M with respect to solute concentrations in solution and 1 atm for gases.

In this case you have the reaction:

3 Cl₂(g) + 2 Fe(s) → 6 Cl⁻(aq) + 2 Fe³⁺(aq)

In this case the following half-reactions occur:

Semi-reaction of oxidation ( an atom or group of atoms loses electrons, or increases its positive charges): Fe³⁺(aq) + 3 e- -->Fe(s); E⁰ = -0.04 V

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In an electrochemical cell at 25°C the potentials of the semi-reactions are usually measured in the sense of reduction and generally the standard potential between both electrochemical cells will be:

$E^{0} =E^{0} _{reduction} -E^{0} _{oxidation}$

E⁰=1.36 V - (-0.04 V)

E⁰=1.36 V + 0.04 V

E⁰=1.40 V

The standard cell potential is 1.40 V. The correct option is the option D (+1.40 V)

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3 years ago

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V2 = volumen of the 80% sugar solution

Equations:
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Sugar from 80% solution: 0.8*V2
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1) 0.5V1 + 0.8V2 = 63

Final volume = volume of 50% solution + volume of 80% solution

2) V1 + V2 = 105

From (2) V1 = 105 - V2

Substitue in (1)

0.5 (105 - V2) + 0.8 V2 = 63

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Answer: 70 mL of the 50% solution and 35 mL of the 80% solution.

5 0

3 years ago