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avanturin [10]
2 years ago
11

Given the values of So given below in J/mol K and the values of ΔHfo given in kJ/mol, calculate the value of ΔGo in kJ for the c

ombustion of 1 mole of propane to form carbon dioxide and gaseous water at 298 K. S (C3H8(g)) = 271 S (O2(g)) = 204 S (CO2(g)) = 214 S (H2O(g)) = 184 ΔHfo (C3H8(g)) = -100 ΔHfo (CO2(g)) = -398 ΔHfo (H2O(g)) = -226
Chemistry
1 answer:
Fantom [35]2 years ago
7 0

Answer:

-2024 kJ

Explanation:

The combustion of propane is

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

And, ΔG° = ΔH° - TΔS°

Where T is the temperature in K.

ΔH° = ∑n*H°f, reactants - ∑n*H°f, products

Where n is the number of moles in the stoichiometry reaction. H°f, O₂(g) = 0 because it's a substance formed by only one element.

ΔH° = [4*(-226) +3*(-398)] - [-100] = -1998 kJ

ΔS° = ∑n*S°, reactants - ∑n*S°, products

ΔS° = [4*(184) + 3*214] - [5*204 + 271] = 87 J/K = 0.087 kJ/K

So

ΔG° = -1998 - 298*0.087

ΔG° = -2024 kJ

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Answer:

true

Explanation:

because the atoms make up the elements which forms the compound

8 0
2 years ago
A Gas OCCUPIES 525ML AT A PRESSURE OF 85.0 kPa WHAT WOULD THE VOLUME OF THE GAS BE AT THE PRESSURE OF 65.0 kPa
german

Boyle's law of ideal gas: This law states that the volume of a gas is inversely proportional to its pressure at a constant temperature. Acc to this law we can write the relation of pressure and volume as:

PV=Constant

That means:

P_{1}V_{1}=P_{2}V_{2}

From that equation we can calculate Volume of gas at a certain pressure:

P₁=Initial pressure

V₁=Initial volume

P₂=Final pressure

V₂= Final volume

Here P₁, initial pressure is given as 85.0 kPa

V₁, initial volume is given as 525 mL

P₂, final pressure is 65.0 kPa

P_{1}V_{1}=P_{2}V_{2}

so,

V_{2}=85\times 525\div 65

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8 0
2 years ago
Read 2 more answers
Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0
ExtremeBDS [4]

Answer:

The freezing point will be 2.9^{0}C

Explanation:

The depression in freezing point is a colligative property.

It is related to molality as:

Depressioninfreezing point=K_{f}Xmolality

Where

Kf= 20\frac{^{0}C}{m}

the molality is calculated as:

molality=\frac{moles_{solute}}{mass_{solvent}}

moles=\frac{mass}{molarmass}=\frac{0.694}{154}=0.0045mol

massofcyclohexane=25g=0.025Kg

molarity=\frac{0.0045}{0.025}=0.18m

Depression in freezing point = 20X0.18=3.6^{0}C

The new freezing point = 6.5^{0}C-3.6^{0}C=2.9^{0}C

5 0
2 years ago
Compound X has the same molecular formula as butane but has a different boiling point and melting point. What can be concluded a
Vsevolod [243]
Compound X is an isomer of butane with different chain types. It is either straight chain or bend chain.
3 0
2 years ago
1. Which liquid sample is a pure substance?
IRISSAK [1]

The whole Activity , poem and paragraph is missing in the question.

Answer:

(1) Liquid A

(2) Solid A

Explanation:

Using this part of the given poem

Substances and mixtures behave differently,

During boiling and melting most especially

Boiling point of substance is fixed while mixture is not

Substance melts completely but mixture does not

The boiling point of the Pure substance remain fixed after reaching its boiling point this is shown by Liquid A

Solid A is melting completely so Solid A is a pure substance.

6 0
2 years ago
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