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iren [92.7K]
3 years ago
7

A chemical reaction is carried out in a closed container. The energy absorbed by the chemical reaction is 50 kJ. What is the ene

rgy liberated from the air or the walls of the container?
A)50 kJ
B)75 kJ
C)25 kJ
D)100 kJ
Chemistry
1 answer:
Contact [7]3 years ago
8 0

Answer:

A) 50 kJ.

Explanation:

  • According to the law of conservation of energy that energy can not be destroyed or created from nothing but it can be converted from one form to another.
  • In a closed system; the amount of heat released by a part of the system is equal to the amount of heat absorbed by other parts of the system.

<em>Q released = Q absorbed.</em>

<u><em>So, the energy liberated from the air or the walls of the container is 50 kJ.</em></u>

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Dalton’s Law CalculationA mixture of H₂, N₂ and Ar gases is present in a steel cylinder. The total pressure within the cylinder
Zolol [24]

Answer:

A) The partial presssure of CO₂ is 167 mm Hg

B) The partial presssure of N₂ is 354 mm Hg

C) The partial presssure of Ar is 235 mm Hg

D) The partial presssure of H₂ is 86 mm Hg

Explanation:

Dalton's law of partial pressures is basically expressed by the following statement:

The total pressure of a mixture is equal to the sum of the partial pressures of its components.

So initially we have:

P_{T}= total presure of the system (675 mm Hg).

P_{N_2}= partial pressure of N₂ (354 mm Hg).

P_{Ar}= partial pressure of Ar (235 mm Hg).

Using Dalton's law we can find the partial pressure of H₂:

P_{T}=P_{N_2}+P_{Ar}+P_{H_2}

675 mm Hg=354 mm Hg + 235 mm Hg + P_{H_2}

P_{H_2}= 675 mm Hg - 354 mm Hg - 235 mm Hg

P_{H_2}=86 mm Hg

If CO₂ gas is added to the mixture, at constant temperature, and the volume is the same, the difference between the new total pressure and the previous total pressure is equal to the partial pressure of CO₂.

P_{T}=P_{N_2}+P_{Ar}+P_{H_2}+P_{CO_2}

842 mm Hg= 354 mm Hg + 235 mm Hg + 86 mm Hg + P_{CO_2}

P_{CO_2}= 842 mm Hg - 354 mm Hg - 235 mm Hg - 86 mm Hg

P_{CO_2}= 167 mm Hg

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Cellulose and starch are examples of:
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Answer:

The choose (d)

d. polysaccharides

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The equilibrium constant for the reaction
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Answer: The concentrations of Cl_2 at equilibrium is 0.023 M

Explanation:

Moles of  Cl_2 = \frac{\text {given mass}}{\text {Molar mass}}=\frac{10g}{71g/mol}=0.14mol

Volume of solution = 1 L

Initial concentration of Cl_2 = \frac{0.14mol}{1L}=0.14M

The given balanced equilibrium reaction is,

                            COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

Initial conc.           0.14 M           0 M       0M    

At eqm. conc.     (0.14-x) M        (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}

Now put all the given values in this expression, we get :

4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}

By solving the term 'x', we get :

x = 0.023 M

Thus, the concentrations of Cl_2 at equilibrium is 0.023 M

7 0
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Natasha and Reanna observe a large airplane in the troposphere. Which experimental setup below would best determine how changing
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Answer:

i am pretty sure the answer is a

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A fly hits a windshield of a truck. The truck exerts a force on the fly, and
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Answer:

A fly hits a windshield of a truck.

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newton's first law

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