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elena-14-01-66 [18.8K]
3 years ago
14

Which of the following reactions show a decrease in entropy? a. CaO(s) + H2O(l) Ca(OH)2 (s)

Chemistry
2 answers:
Mariulka [41]3 years ago
7 0
The correct answer is <span>a. CaO(s) + H2O(l) Ca(OH)2 (s)
</span><span>Decreasing in entropy means that you have to get a solid from the liquid
</span>therefore : <span>CaO(s) + H2O(l) Ca(OH)2 (s)
This formula will make it absolutely clear 
</span>(g) -\ \textgreater \ (l) -\ \textgreater \ (s) more molecules -\ \textgreater \  less molecules&#10;
muminat3 years ago
4 0

<u>Answer:</u> The correct answer is Option a.

<u>Explanation:</u>

Entropy is defined as the measurement of randomness in a system.

Entropy increases as we move from solid state to liquid state to gaseous state and it decreases as we move from gaseous state to liquid state to solid state.

From the options above:

<u>Option a:</u> CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)

Entropy is getting decreased because liquid reactants are changing its state to solid state.

<u>Option b:</u> 2H_2O(l)\rightarrow 2H_2(g)+O_2(g)

In this reaction, entropy is increasing because the state is changing from liquid to gaseous state.

<u>Option c: </u>H_2CO_3(aq.)\rightarrow CO_2(g)+H_2O(l)

In this reaction, entropy is increasing because the state is changing from liquid to gaseous state.

<u>Option d:</u> 2KClO_3(s)\rightarrow 3O_2(g)+2KCl(s)

In this reaction, entropy is increasing because the state is changing from solid to gaseous state.

Hence, the correct answer is Option a.

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Using henry's law, calculate the molar concentration of o2 in the surface water of a mountain lake saturated with air at 20 ∘c a
s2008m [1.1K]

Answer: The molar concentration of oxygen gas in water is 1.43\times 10^{-7} mol/L.

Explanation:

Partial pressure of the O_2gas = 685 torr = 0.8905 bar

1 torr = 0.0013 bar

According Henry's law:

p_{o_2}=K_H\times\chi_{O_2}

Value of Henry's constant of oxygen gas at 20 °C in water = 34860 bar

0.0013=34860 bar\times \chi_{O_2}

\chi_{O_2}=\frac{0.8905 bar}{34680 bar}=2.56\times 10^{-5}

Let the number of moles of O_2 gas in 1 liter water be n.

1 Liter water = 1000 g of water

Moles of water in 1 L n_w=\frac{1000 g}{18 g/mol}=55.55 mol

\chi_{O_2}=\frac{n}{n+n_w}

2.56\times 10^{-5}=\frac{n}{n+55.55}

n=1.43\times 10^{-7} moles

Molarity=\frac{\text{Moles of}O_2}{Volume}

Molar concentration of oxygen gas in 1 L of water:

=\frac{1.43\times 10^{-7} moles}{1 L}=1.43\times 10^{-7} mol/L

The molar concentration of oxygen gas in water is 1.43\times 10^{-7} mol/L.

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