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oksano4ka [1.4K]
2 years ago
5

Do you feel that this equation is currently following the law of conservation of mass?

Chemistry
1 answer:
mart [117]2 years ago
7 0

Answer:

N2 + H2 --> 2 NH

Explanation:

You need to add 2 to make it correct

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D. the can from the car because there are fewer solute-solvent collisions.
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Entropy of a system decreases with decrease in Temperature T/F
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Answer: The given statement is true.

Explanation:

Entropy means the measure of randomness present in a substance. That is, an increase in temperature will lead cause more motion in the particles of a substance more will be their kinetic energy.

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So, when we decrease the temperature then there will be decrease in motion of particles. As a result, lesser number of collisions will take place between them. Hence, degree of randomness will also decrease.

Thus, we can conclude the statement entropy of a system decreases with decrease in temperature, is true.

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When air bags inflate, nitrogen gas is formed from sodium azide, along with solid sodium. what reaction category is this?
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3 years ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
Help pls I will mark brainlest​
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Resources found in lithosphere: gold and iron etc

Resources found in atmosphere: Water vapor, gases etc.

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