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user100 [1]
4 years ago
14

In which of these systems is the entropy decreasing?

Chemistry
2 answers:
atroni [7]4 years ago
6 0

I think it is aaaaaaaa I think it's aaa

DochEvi [55]4 years ago
5 0
<h2>Answer:</h2>

The entropy decreases in <u>B. a gas condensing</u>.

<h2>Explanation:</h2>

Entropy refers to phenomenon which measures the spreading of energy in a system. The energy of the system is spreading it has high entropy, whereas if the energy of the system remains concentrated it is said to have decreased entropy.

In case of air escaping from the tire the energy is spreading out that's why the air is moving out of the tire. Thus, it has high entropy. In case of gas condensing which leads to the formation of liquid, the energy of the gaseous particles remains concentrated such that it may lead to the formation of liquid. Thus, this system has decreased entropy.

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Wich statement about nuclear binding energy is true
goldfiish [28.3K]
Can u plz tell where are the statements
7 0
4 years ago
Which metal is more active Ni and less active than Zn
arlik [135]
Is this a multiple choice?
3 0
3 years ago
How to solve for K when given your anode and cathode equations and voltage
Aleks04 [339]

Answer:

See Explanation

Explanation:

In thermodynamics theory the Free Energy (ΔG) of a chemical system is described by the expression ΔG = ΔG° + RTlnQ. When chemical system is at equilibrium ΔG = 0. Substituting into the system expression gives ...

0 = ΔG° + RTlnKc, which rearranges to ΔG° = - RTlnKc.  ΔG° in electrochemical terms gives ΔG° = - nFE°, where n = charge transfer, F = Faraday Constant = 96,500 amp·sec and E° = Standard Reduction Potential of the electrochemical system of interest.

Substituting into the ΔG° expression above gives

-nFE°(cell) = -RTlnKc => E°(cell) = (-RT/-nF)lnKc = (2.303·R·T/n·F)logKc

=> E°(cell) = (0.0592/n)logKc = E°(Reduction) - E°(Oxidation)

Application example:

Calculate the Kc value for a Zinc/Copper electrochemical cell.

Zn° => Zn⁺² + 2e⁻  ;    E°(Zn) = -0.76 volt  

Cu° => Cu⁺² + 2e⁻ ;    E°(Cu) =  0.34 volt

By natural process, charge transfer occurs from the more negative reduction potential to the more positive reduction potential.

That is,

           Zn° => Zn⁺² + 2e⁻ (Oxidation Rxn)

Cu⁺² + 2e⁻ => Cu°             (Reduction Rxn)

E°(Zn/Cu) = (0.0592/n)logKc

= (0.0592/2)logKc = E°(Cu) - E°(Zn) = 0.34v - (-0.76v) = 1.10v

=> logKc = 2(1.10)/0.0592 = 37.2

=> Kc = 10³⁷°² = 1.45 x 10³⁷

3 0
3 years ago
Distinguish between single and double replacement reactions and give an example of<br> each.
Alenkasestr [34]

Answer:

i think this is right!

Explanation:

An example of a single replacement reaction occurs when potassium (K) reacts with water (H2O). A colorless solid compound named potassium hydroxide (KOH) forms, and hydrogen gas (H2) is set free. The equation for the reaction is: 2K + 2H2O → 2KOH + H.

An example of a double replacement reaction is the reaction between silver nitrate and sodium chloride in water. Both silver nitrate and sodium chloride are ionic compounds. Both reactants dissolve into their ions in aqueous solution.

6 0
2 years ago
A gas sample is held at constant pressure. The gas occupies 2.97 L of volume when the temperature is 21.6°C. Determine the tempe
STALIN [3.7K]

Answer:

339.2K

Explanation:

Using Charles law equation;

V1/T1 = V2/T2

Where;

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

V1 = 2.97 L

V2 = 3.42 L

T1 = 21.6°C = 21.6 + 273 = 294.6K

T2 = ?

Using V1/T1 = V2/T2

2.97/294.6 = 3.42/T2

Cross multiply

2.97 × T2 = 294.6 × 3.42

2.97T2 = 1007.532

T2 = 1007.532 ÷ 2.97

T2 = 339.236

The final temperature is 339.2K

5 0
3 years ago
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