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yarga [219]
2 years ago
13

A potential energy diagram is shown.

Chemistry
1 answer:
Vesna [10]2 years ago
5 0

Answer:

25kJ

Explanation:

Given the initial energy to be 30kJ

The energy change from the initial energy to the peak energy = (65-30) kJ

= 35kJ

since the second energy change was a drop in energy it is regarded negative

= (55-65)

= -10kJ

Therefore total energy change

= (35-10)kJ

= 25kJ

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ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J.
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Complete question:

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.

Answer:

The magnitude of q for the process 568 J.

Explanation:

Given;

change in internal energy of the gas, ΔU = 475 J

work done by the gas, w = 93 J

heat added to the system, = q

During gas expansion process, heat is added to the gas.

Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.

ΔU = q - w

q = ΔU +  w

q = 475 J  +  93 J

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What is the percent by mass of Fluorine in Nitrogen triflouride NF3?
Kipish [7]

Answer:

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Explanation:

Data

Percent by mass of F

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Process

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2.- Use proportions and cross multiplications to find the percent by mass of F. The molar mass of NF₃ is equal to 100%.

                       71 g of NF₃ ------------------ 100%

                       57 g of F   ------------------- x

                            x = (57 x 100)/71

                            x = 5700 / 71

                            x = 80.3%

3.- Conclusion

Fluorine is 80.3% by mass of the molecule NF₃

           

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