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Annette [7]
3 years ago
7

A reaction begins with 1.00 mol of HCl in a 4.00 liter container at a certain temperature.

Chemistry
1 answer:
lubasha [3.4K]3 years ago
3 0

Answer:

[HCl]_{eq}=0.05M

Explanation:

Hello,

In this case, for the given chemical reaction, the law of mass action at equilibrium results:

Kc=\frac{[H_2]_{eq}[Cl_2]_{eq}}{[HCl]^2_{eq}}

Next, in terms of the change x due to reaction extent, it is rewritten, considering an initial concentration of HCl of 0.25M (1mol/4L), as:

4.00=\frac{(x)(x)}{(0.25-2x)^2}

Thus, solving for x via quadratic equation or solver, the following results are obtained:

x_1=0.1M\\x_2=0.17M

Clearly, the solution is x=0.1M as the other result will provide a negative concentration for the hydrochloric acid at equilibrium, thereby, its equilibrium concentration turns out:

[HCl]_{eq}=0.25M-2*0.1M

[HCl]_{eq}=0.05M

Best regards.

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Explanation:

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6 0
2 years ago
10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C
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Answer:

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Explanation:

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c = specific heat of the substance

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1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c =  0.385 J/g°C

\Delta T=\frac{Q}{mc}

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2) 10.0 g of aluminium

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m = 10.0 g

Specific heat of the aluminium= c =  0.903 J/g°C

\Delta T=\frac{Q}{mc}

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3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c =  2.42 J/g°C

\Delta T=\frac{Q}{mc}

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4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c =  4.18J/g°C

\Delta T=\frac{Q}{mc}

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Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c =  0.128 J/g°C

\Delta T=\frac{Q}{mc}

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Hope this helps!!
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