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olga nikolaevna [1]
3 years ago
11

What does the second law of thermodynamics say about entropy

Chemistry
2 answers:
zzz [600]3 years ago
7 0

Answer: The entropy in a system and its surroundings tends to increase.

Explanation:

3.3.2 Quiz: Entropy (a p e x)

alukav5142 [94]3 years ago
5 0

Answer:

The entropy and the systems surrounding it tend to increase.

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2. What happens when hydrochloric acid (HCl) is added to the solution? Do the relative concentrations of H+, CH3COOH, or
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Answer: A molecule of hydrochloric acid, for example, is composed of a hydrogen atom and a chlorine atom. When these molecules dissolve into water, they separate into a positively charged hydrogen ion and a negatively charged chlorine ion. ... Only some of the molecules of weak acids disassociate when added to water.

Explanation:

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3 years ago
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In addition to the two types of motion found in liquids, gases have <br> What kind of motion
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Answer:

In liquids, particles are quite close together and move with random motion throughout the container. Particles move rapidly in all directions but collide with each other more frequently than in gases due to shorter distances between particles.

3 0
3 years ago
Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) → 2SO3(g) Suppose the volume of this system is c
oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be  increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen  (i.e 2 moles +1 mole  =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be P_1

and the total pressure at the final equilibrium be P_2

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝  1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

P_1V_1 = P_2V_2

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

V_2 =  \dfrac{V_1}{\dfrac{3}{2}} ----- (1)

also;

Thus ;

P_1V_1 = P_2(  \dfrac{V_1}{\frac{3}{2}})

P_1V_1 = P_2 * 2  \dfrac{V_1}{3}

3 P_1 V_1 = 2 P_2 V_1

Dividing both sides by V_1

3P_1 = 2P_2

P_2 =P_1 \dfrac{3}{2}  ----- (2)

From ;

P_1V_1 = P_2V_2

P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{V_1}{\frac{3}{2}}

P_2 V_2 = P_1 * \dfrac{3}{2}*   \dfrac{2 }{3}}*V_1

P_2 V_2 = P_1 V_1

Thus; The new equilibrium total pressure will be  increased to one-half to initial total pressure.

7 0
3 years ago
Just need a letter so no that does not help
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Like A B C D or like a mailed letter

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What is the democritus findings?
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