The entropy of a substance above absolute zero will always be: a. Negative
b. Positive
c. Neither Negative nor positive

Answers

Answer 1

Answer: i will be positive. just because it’s positive

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Partc. explain why equal volumes of 0.1 m ch3cooh and 0.1 m nach3co2 function as a buffer solution, but equal volumes of 0.1 m hcl and 0.1 m naoh do not.

Answers

Answer : As per the definition of the buffer solution "It is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa".

so considering the above given condition for $CH_(3) COOH$ and $NaCH_(3)CO_(2)$ act as buffer because it satisfies and qualifies the conditions of behaving as a buffer when mixed in equal amounts,

whereas HCl and NaOH are strong acid and strong base which does not satisfies the criteria for being called as buffer.

As buffer needs a weak acid and its conjugate base to behave and act as buffer and resist any pH change.

Assuming 100% dissociation, which of the following compounds is listed incorrectly with its van't Hoff factor i? Al2(SO4)3, i = 4 NH4NO3, i = 2 Mg(NO3)2, i = 3 Na2SO4, i = 3 Sucrose, i = 1

Answers

Answer:

- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).

Explanation:

Hello,

In this case, since the van't Hoff factor is related with the species that result from the ionization of a chemical compound, we can see that that

- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).

- Ammonium nitrate NH4NO3 dissociates in one ammonium ions and one nitrate ion, therefore, van't Hoff factor is 2 (correct).

- Sodium sulfate Na2SO4 dissociates in two sodium ions and one sulfate, therefore, van't Hoff factor is 3 (correct).

- Sucrose is not ionized, therefore, van't Hoff factor is 1 (correct).

Best regards.

What is the initial temperature of a gas if the volume changed from 1.00l to 1.10l and the final temperature was determined to be 255?

Answers

Answer is 231.8 K

Explanation
We can use Charle's law to solve this problem.

Charle's law says"at a constantpressure, the volume of a fixed amount of gas is directly proportional to itsabsolute temperature".

V α T

Where V is the volume and T is the temperaturein Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂

V₁ = 1.00 L
T₁ =?
V₂ = 1.10 L
T₂ = 255 K (Since the unit of the given temperature is missing in the question, thought that the given temperature is in Kelvin)

By applying theformula,
1.00 L / T₁ = 1.10 L / 255 K
T₁ =(1.00 L / 1.10 L) x 255 K
T₁ = 231.8 K

Hence, the initial temperature is 231.8 K.

Assumption : the pressure of the gas is a constant.

Hello!

What is the initial temperature of a gas if the volume changed from 1.00l to 1.10l and the final temperature was determined to be 255 ?

We have the following data:

V1 (initial volume) = 1.00 L

V2 (final volume) = 1.10 L

T1 (initial temperature) = ? (in Kelvin)

T2 (final temperature) = 255 K

According to the Law of Charles and Gay-Lussac in the study of gases, in an isobaric transformation, ie when a mass under pressure maintains its constant pressure, on the other hand, as the volume increases, the temperature increases and, if the volume decreases, the temperature decreases (directly proportional to temperature and volume) . We apply the data to the formula of isobaric transformation (Charles and Gay-Lussac), we will see:

$(V_1)/(T_1) = (V_2)/(T_2)$

$(1.00)/(T_1) = (1.10)/(255)$

multiply the means by the extremes

$1.10*T_1 = 1.00*255$

$1.10\:T_1 = 255$

$T_1 = (255)/(1.10)$

$\boxed{\boxed{T_1 \approx 231.8\:K}}\:\:\:\:\:\:\bf\red{\checkmark}$

Answer:

The initial temperature is approximately 231.8 Kelvin

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The destruction of the ozone layer by chlorofluorocarbons (CFC’s) can be described by the following reactions:ClO(g) + O3(g) ? Cl(g) + 2 O2(g) ?H°rxn = –29.90 kJ2 O3(g) ? 3 O2(g) ?H°rxn = 24.18 kJDetermine the value of heat of reaction for the following:Cl(g) + O3(g) ? ClO(g) + O2(g) ?H°=_____________?

Answers

Answer:

ΔH°rxn = 54.08 kJ

Explanation:

Let's consider the following equations.

a) ClO(g) + O₃(g) ⇄ Cl(g) + 2 O₂(g) ΔH°rxn = –29.90 kJ

b) 2 O₃(g) ⇄ 3 O₂(g) ΔH°rxn = 24.18 kJ

We have to determine the value of heat of reaction for the following reaction: Cl(g) + O₃(g) ⇄ ClO(g) + O₂(g)

According to Hess's law, the enthalpy change in a chemical reaction is the same whether the reaction takes place in one or in several steps. That means that we can find the enthalpy of a reaction by adding the corresponding steps and adding their enthalpies. According to Lavoisier-Laplace's law, if we reverse a reaction, we also have to reverse the sign of its enthalpy.

Let's reverse equation a) and add it to equation b).

-a) Cl(g) + 2 O₂(g) ⇄ ClO(g) + O₃(g) ΔH°rxn = 29.90 kJ

b) 2 O₃(g) ⇄ 3 O₂(g) ΔH°rxn = 24.18 kJ

-------------------------------------------------------------------------------------------------

Cl(g) + 2 O₂(g) + 2 O₃(g) ⇄ ClO(g) + O₃(g) + 3 O₂(g)

Cl(g) + O₃(g) ⇄ ClO(g) +O₂(g)

ΔH°rxn = 29.90 kJ + 24.18 kJ = 54.08 kJ

Final answer:

The heat of the reaction (ΔH°rxn) for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g) is calculated using Hess's Law. The sum of the heat of reversed first reaction and the second reaction provided is 54.08 kJ.

Explanation:

The chemistry question asks to determine the heat of the reaction for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g). In Hess's Law, the heat of the reaction or ΔH for a reaction can be calculated from the sum of the heats of other reactions that sum to the desired reaction. In this case, we want to reverse the first reaction provided (which changes the sign of ΔH) and add it to the second reaction provided.

So, reversing the first reaction we get: Cl(g) + 2 O2(g) ? ClO(g) + O3(g) ?H°rxn = 29.90 kJ

Adding this to the second reaction: 2 O3(g) ? 3 O2(g), ?H°rxn = 24.18 kJ, gives the reaction Cl(g) + O3(g) ? ClO(g) + O2(g). Adding the ΔH values gives the ΔH for this reaction: 29.90 kJ + 24.18 kJ = 54.08 kJ. So, ?H°rxn for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g) is 54.08 kJ.

Learn more about Heat of Reaction here:

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Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mLmL of water to produce a solution that freezes at −−14.5 ∘C? The freezing point for pure water is 0.0 ∘C∘C and Kf is equal to 1.86 ∘C/m.If the 3.90 m solution from Part A boils at 103.45 ∘C, what is the actual value of the van't Hoff factor, i? The boiling point of pure water is 100.00 ∘C and Kb is equal to 0.512 ∘C/m.

Answers

Answer:

1) 108.27 grams of potassium nitarte must be added to 275 mL of water to produce a solution that freezes at -14.5°C.

2) 1.73 is the actual value of the van't Hoff factor, i.

Explanation:

1) Formula used depression in freezing point ;

$\Delta T_f=T-T_f$

$\Delta T_f=i* K_f* m$

where,

$T_f$ =Freezing point of solution

T = Freezing point of water

$\Delta T_f$ =depression in freezing point =

i = van't Hoff factor of solute

$K_f$ = freezing point constant

m = molality of solution

We have :

$K_f$ of water = 1.86°C/m ,

Molality of solution = m = ?

$KNO_3(aq)\rightarrow K^+(aq)+NO_3^(-)(aq)$

i = 2

Freezing point of solution = $T_f=-14.5^oC$

Freezing point of pure water = T = 0°C

$\Delta T_f=T-T_f$

$\Delta T_f=0^oC-(-14.5 ^oC)=14.5^oC$

$14.5^oC=2* 1.86^oC* m$

m = 3.898 molal

3.898 moles of potassium nitrate is dissolved in 1 kg of water or 1000 g of water.

Volume of water , V= 275 ml

Mass of water = m

Density of water= d = 1 g/mL

$m=d* V=1 g/ml* 275 mL = 275 g$

Here, 3.898 moles of potassium nitrate is dissolved in 1 kg of water or 1000 g of water. Then moles of potassium nitarte present in 275 grams of water is :

$(3.989 mol)/(1000)* 275 =1.072 mol$

Mass of 1.072 moles of potassium nitrate :

1.072 mol × 101 g/mol = 108.27 g

108.27 grams of potassium nitarte must be added to 275 mL of water to produce a solution that freezes at -14.5°C.

2) Formula used an Elevation in boiling point;

$\Delta T_b=T_b-T$

$\Delta T_b=i* K_b* m$

where,

$T_b$ =boiling point of solution

T = boiling point of water

$\Delta T_b$ =Elevation in boiling point =

i = van't Hoff factor of solute

$K_b$ = Boiling point constant

m = molality of solution

of the solution

We have :

$K_b$ of water = 0.512°C/m ,

Molality of solution = m = 3.90 m

i =?

The boiling point of pure water = T = 100.00°C

The boiling point of solution = $T_b$ = 103.45°C

$\Delta T_b=103.45^oC-100.00^oC=3.45^oC$

$\Delta T_b=i* K_b* m$

$3.45^oC=i* 0.512 ^oC/m* 3.90 m$

$i=(3.45^oC)/(0.512 ^oC/m* 3.90 m)=1.73$

1.73 is the actual value of the van't Hoff factor, i.

.......state Hess law

Answers

Hess's Law of Constant Heat Summation (or just Hess's Law) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.

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