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Lapatulllka [165]
3 years ago
15

Which of the following processes have a ?S > 0?

Chemistry
1 answer:
Ne4ueva [31]3 years ago
3 0

Answer: Cesium chloride dissolves in pure water has \Delta S > 0.

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) 2HCl(g)\rightarrow H_2(g)+Cl_2(l)

As 2 moles of gas are converting to 1 mole of gas and 1 mole of liquid, the randomness decreases and hence \Delta S is negative.

b) 2NH_3(g)+CO_2(g)\rightarrow NH_2CONH_2(aq)+H_2O(l)

As gas molecules are converting into aqueous and liquid form,  the randomness decreases and hence \Delta S is negative.

c) lithium fluoride forms from its elements: 2Li+F_2\rightarrow 2LiF

As elements are combining to form a compound, the randomness decreases and hence \Delta S is negative.

d) Cesium chloride dissolves in pure water : CaCl_2(s)\rightarrow Ca^{2+}(aq)+2Cl^-(aq)

As compound is dissociating into ions, the randomness increases and hence \Delta S is positive.

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Answer:

B. temperature decreases as altitude increases.

Explanation:

Just like in the lower reaches of the atmosphere, the troposphere, in the mesosphere, temperature decreases as altitude increases.

The mesosphere is the third layer of the atmosphere just above the stratosphere.

  • It begins at the top of the stratosphere and ends at the mesopause where the thermosphere begins.
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With increasing height, the temperature of the mesosphere decreases significantly. The top of the mesosphere is one of the coldest part of the earth atmosphere. This is as a result of increasing atmospheric cooling by carbon dioxide in this region of the atmosphere.

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3 years ago
If the freezing point of the solution had been incorrectly read 0.3 °C lower than the true freezing point, would the calculated
Dovator [93]

Answer : The molar mass of the solute would be low.

Explanation :

Formula used for depression in freezing point is:  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{w_b}{M_b}\times w_a}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution

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i = Van't Hoff factor

K_f = freezing point constant

m = molality

w_b = mass of solute

w_a = mass of solvent

M_b = molar mass of solute

From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.

Hence, the molar mass of the solute would be low.

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