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arsen [322]
3 years ago
8

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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jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

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Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

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The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

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The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

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According to the given reaction:

K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate  = 0.004 moles

Available moles of barium nitrate  =  0.012 moles

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The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

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0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

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<u>(B) is correct.</u>

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