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Fofino [41]
3 years ago
13

Which one of the following processes produces a decrease in the entropy of the system?a, precipitation of AgCl(s) from Ag+(aq) a

nd Cl−(aq) ions in solutionb. dissolution of LiOH(s) in waterc. melting solid gold into liquid goldd. evaporation of Hg(l) to form Hg(g)e. mixing of two gases into one container SubmitGive Up Continue
Chemistry
1 answer:
Usimov [2.4K]3 years ago
4 0

Answer:  precipitation of AgCl(s) from Ag^+(aq) and Cl^-(aq) ions in solution

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

\Delta S is positive when randomness increases and \Delta S is negative when randomness decreases.

a) precipitation of AgCl(s) from Ag^+(aq) and Cl^-(aq) ions in solution : As ions are getting solidified, entropy decreases.

b) dissolution of LiOH(s) in water: The compounds dissociates into ions, entropy increases.

c) melting solid gold into liquid gold: The randomness increases, entropy increases.

d) evaporation of Hg(l) to form Hg(g) : The randomness increases, entropy increases.

e)  mixing of two gases into one container : The randomness increases, entropy increases.

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Explanation:

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3 years ago
G you have 57.0 ml of a 0.400 m stock solution that must be diluted to 0.100 m. assuming the volumes are additive, how much wate
Amanda [17]
Answer:
added water = 171 ml

Explanation:
Assuming volumes are additive, the rule that we will use to solve this question is:
M1V1 = M2V2
where:
M1 is the initial concentration = 0.4 m
V1 is the initial volume = 57 ml
M2 is the final concentration = 0.1 m
V2 is the final volume that we want to calculate
Substitute with the given in the above equation to get V2 as follows:
M1V1 = M2V2
(0.4)(57) = (0.1)V2
22.8 = 0.1V2
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Now, the final volume is equal to the initial volume plus the amount of added water. So, to get the amount of added water, we will subtract the initial volume from the final volume as follows:
V2 = V1 + added water
228 = 57 + added water
added water = 228 - 57 = 171 ml

Hope this helps :)
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number of moles:

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M = moles / Volume ( in liters )

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M = 0.1760 mol/L⁻¹

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