Which property of a substance is not altered by a physical change?

For each process, predict the sign on the entropy change and write a sentence or two to explain how to make this prediction with

MrRa [10]

Answer: a. A solid melts: $\Delta S=+ve$

b. a vapor is converted to solid: $\Delta S=-ve$

c. a liquid freezes: $\Delta S=-ve$

d. A solid sublimes: $\Delta S=+ve$

e. a vapor condenses to liquid : $\Delta S=-ve$

f. a liquid boils: $\Delta S=+ve$

g. dissolving a tablespoon of salt in water: $\Delta S=+ve$

h. combustion of gasoline: $\Delta S=+ve$

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 change is negative and vice versa.

a. A solid melts: The solid is converting to liquid, thus the randomness is increasing as the molecules are moving away and the intermolecular forces are getting weaker. Thus the entropy change $(\Delta S)$ is positive.

b. a vapor is converted to solid: The gas is converting to solid, thus the randomness is decreasing as the molecules are moving close and the intermolecular forces are getting stronger Thus the entropy change $(\Delta S)$ is negative.

c. a liquid freezes: The liquid is converting to solid, thus the randomness is decreasing as the molecules are moving close and the intermolecular forces are getting stronger. Thus the entropy change $(\Delta S)$ is negative.

d. A solid sublimes: The solid is converting to gas, thus the randomness is increasing as the molecules are moving away and the intermolecular forces are getting weaker. Thus the entropy change $(\Delta S)$ is positive.

e. a vapor condenses to liquid : The gas is converting to liquid, thus the randomness is decreasing as the molecules are moving close and the intermolecular forces are getting stronger. Thus the entropy change $(\Delta S)$ is negative.

f. a liquid boils: The liquid is converting to gas, thus the randomness is increasing as the molecules are moving away and the intermolecular forces are getting weaker. Thus the entropy change $(\Delta S)$ is positive.

g. dissolving a tablespoon of salt in water: The solid is converting to ions , thus the randomness is increasing as the ions can move freely. Thus the entropy change $(\Delta S)$ is positive.

h. combustion of gasoline: The liquid is converting to gas, thus the randomness is increasing as the molecules are moving away and the intermolecular forces are getting weaker. Thus the entropy change $(\Delta S)$ is positive.

6 0

2 years ago

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

$CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

$CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

$[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$

= $0.032 M/atm \times 1.9 \times 10^{-4}atm$

= $0.0608 \times 10^{-4}$

or, = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As, $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

$4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$

$[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

$[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So, pH = $-log (1.64 \times 10^{-6})$

= 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0

3 years ago

3. Calculate the answers to the appropriate number of significant figures. e) 43.678 x 64.1 = f) 1.678/0.42 =

Sophie [7]

Where’s the pictures

8 0

3 years ago

When 551. mg of a certain molecular compound X are dissolved in 100 g of benzonitrile (CH,CN), the freezing point of the solutio

arsen [322]

Answer:

1.12g/mol

Explanation:

The freezing point depression of a solvent for the addition of a solute follows the equation:

ΔT = Kf*m*i

Where ΔT is change in temperature (Benzonitrile freezing point: -12.82°C; Freezing point solution: 13.4°C)

ΔT = 13.4°C - (-12.82) = 26.22°C

m is molality of the solution

Kf is freezing point depression constant of benzonitrile (5.35°Ckgmol⁻¹)

And i is Van't Hoff factor (1 for all solutes in benzonitrile)

Replacing:

26.22°C = 5.35°Ckgmol⁻¹*m*1

4.90mol/kg = molality of the compound X

As the mass of the solvent is 100g = 0.100kg:

4.9mol/kg * 0.100kg = 0.490moles

There are 0.490 moles of X in 551mg = 0.551g, the molar mass (Ratio of grams and moles) is:

0.551g / 0.490mol

= 1.12g/mol

This result has no sense but is the result by using the freezing point of the solution = 13.4°C. Has more sense a value of -13.4°C.

5 0

2 years ago

How many grams of HF are needed to react with 3.0 moles of Sn? *

Lostsunrise [7]

Answer:

120g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

Sn + 2HF —> SnF2 + H2

Next, we shall determine the number of mole of HF needed to react with 3 moles of Sn.

From the balanced equation above, 1 mole of Sn reacted with 2 moles of HF.

Therefore, 3 moles of Sn will react with = 3 x 2 = 6 moles of HF.

Finally, we shall convert 6moles of HF to grams

This is illustrated below:

Number of mole of HF = 6moles

Molar Mass of HF = 1 + 19 = 20g/mol

Mass of HF =..?

Mass = number of mole x molar Mass

Mass of HF = 6 x 20

Mass of HF = 120g

Therefore, 120g of HF is needed to react with 3 moles of Sn

6 0

3 years ago