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2 years ago

The value of delta G at 141.0 degrees celsius for the formation of phosphorous trichloride from its constituent elements,

Chemistry

1 answer:

AURORKA [14]2 years ago

6 0

Answer:

The correct answer is option E.

Explanation:

The Gibbs free energy is given by expression:

ΔG = ΔH - TΔS

ΔH = Enthalpy change of the reaction

T = Temperature of the reaction

ΔS = Entropy change

We have :

ΔH = -720.5 kJ/mol = -720500 J/mol (1 kJ = 1000 J)

ΔS = -263.7 J/K

T = 141.0°C = 414.15 K

$\Delta G = -720500 J/mol - (414.15 K\times (-263.7 J/K))$

$= -611,288.64 J/mol = -611.28 kJ/mol\approx -611.3 kJ/mol$

The Gibb's free energy of the given reaction at 141.0°C is -611.3 kJ/mol.

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joja [24]

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1 year ago

Water is dissolved into n-butanol (a polar liquid). Which is the second step at the molecular level as water dissolves into n-bu

hodyreva [135]

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2 years ago

Read 2 more answers

The molar heat capacity of an unknown substance is 92.1 J/mol-K. If the unknown has a molar mass of 118 g/mol, what is the speci

dlinn [17]

Answer : The specific heat (J/g-K) of this substance is, 0.780 J/g.K

Explanation :

Molar heat capacity : It is defined as the amount of heat absorbed by one mole of a substance to raise its temperature by one degree Celsius.

1 mole of substance releases heat = 92.1 J/K

As we are given, molar mass of unknown substance is, 118 g/mol that means, the mass of 1 mole of substance is, 118 g.

As, 118 g of substance releases heat = 92.1 J/K

So, 1 g of substance releases heat = $\frac{92.1}{118}=0.780J/g.K$

Thus, the specific heat (J/g-K) of this substance is, 0.780 J/g.K

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2 years ago

A 250. ml sample of 0.0328m hcl is partially neutralized by the addition of 100. ml of 0.0245m naoh. find the concentration of h

Ksenya-84 [330]

Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.

Explanation:

1) Molarity of 0.250 L HCl solution : 0.0328 M

$Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0328=\frac{\text{Number of moles}}{0.250 L}$

Moles of HCl in 0.250 L solution = 0.0082 moles

2) Molarity of 0.100 L NaOH solution : 0.0245 M

$Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0245=\frac{\text{Number of moles}}{0.100 L}$

Moles of NaOH in 0.100 L solution = 0.00245 moles

3) Concentration of hydrochloric acid in the resulting solution.

0.00245 moles of NaOH will neutralize 0.00245 moles of HCl out of 0.0082 moles of HCl.

Now the new volume of the solution = 0.100 L +0.250 L = 0.350 L

Moles of HCl left un-neutralized = 0.0082 moles - 0.00245 moles = 0.00575 moles

$Molarity=\frac{\text{number of moles}}{\text{volume of solution in L}}$

Molarity of HCl left un-neutralized : $\frac{0.00575 moles}{0.350L}=0.0164 M$

0.0164 molar concentration of hydrochloric acid in the resulting solution.

3 0

2 years ago