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WARRIOR [948]
3 years ago
10

If under a given set of conditions the reaction A → B occurs with ΔG = -14 kJ/mol, and the reaction C→ B occurs with ΔG =- 16 kJ

/mol, then:_______ a. Conversion of A to C is exergonic. b. A and C can never be at equilibrium, even under different reaction conditions. c. Oconversion of A to C is entropically driven. d. Conversion of C to A is freely reversible.
Chemistry
1 answer:
Inessa [10]3 years ago
4 0

Answer:

a. Conversion of A to C is exergonic.

Explanation:

In the problem:  C→ B occurs with ΔG = + 16 kJ/mol

It is possible to sum ΔG of reactions to obtain ΔG of another related reaction:

A → B ΔG = -14kJ/mol

C → B ΔG = +16kJ/mol

A → C ΔG = -14kJ/mol - (+16kJ/mol) = -30kJ/mol

As ΔG < 0

The reaction is exergonic

Under these condition, the reaction occurs. But under another conditions, the reaction will be at equilibrium (ΔG = 0)

A reaction is entropically driven if ΔG < 0 and ΔS is high. But we don't have information of ΔS.

As ΔG <0, the reaction is not spontaneous in the reverse direction

Right option is:

<h3>a. Conversion of A to C is exergonic.</h3>

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A 58.33-g sample of milk of magnesia, Mg(OH)2, always contains 24.31g of magnesium, 32.00g of oxygen, and 2.02g of hydrogen. Fin
ASHA 777 [7]

The mass percentage of each element in the milk of magnesia is, 41.68% Mg, 54.86% Oxygen, and 3.46% Hydrogen elements

Give the total mass of a sample of milk of magnesia Mg(OH)₂ as 58.33 g

We are to calculate the percentage by mass of each element

\% mass =\frac{mass \ of \ element}{mass \ of \ sample} \times 100

For Magnesium:

Mass of magnesium = 24.31g

\% mass \ of \ Mg = \frac{24.31}{58.33} \times 100\\ \% mass \ of \ Mg =\frac{2431}{58.33} \\ \% mass \ of \ Mg = 41.68\%

For oxygen element:

Mass of oxygen = 32.00g

\% mass \ of \ O_2 = \frac{32}{58.33} \times 100\\ \% mass \ of \ O_2 =\frac{3200}{58.33} \\ \% mass \ of \ O_2 = 54.86\%

For the hydrogen element

Mass of hydrogen = 2.02g

\% mass \ of \ H_2 = \frac{2.02}{58.33} \times 100\\ \% mass \ of \ H_2 =\frac{202}{58.33} \\ \% mass \ of \ H_2 = 3.46\%

Hence the mass percentage of each element in the milk of magnesia is, 41.68% Mg, 54.86% Oxygen, and 3.46% Hydrogen elements.

Learn more here: brainly.com/question/20065048

5 0
3 years ago
Two mice have different parents. They both have the same protein for whisker thickness in their cells.
musickatia [10]

Answer:

Due to presence of similar protein.

Explanation:

The two mice are different from one another due to their different parents so in that way the physical features of both mice are different from one another. Both mice have the same protein for whisker thickness in their cells that leads to same whisker in both mice. The common feature in both mice are the protein that is responsible for whisker thickness so both have same whisker on their face.

3 0
3 years ago
A steel container filled with H₂ gas is at a pressure of 6.5 atm and a temperature of 22°C. If the container is placed near a fu
Yuki888 [10]

Answer: The new pressure is 7.1 atm

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=6.5atm\\T_1=22^0C=(22+273)K=295K\\P_2=?\\T_2=50^0C=(50+273)K=323K

Putting values in above equation, we get:

\frac{6.5}{295}=\frac{P_2}{323}\\\\P_2=7.1

Hence, the new pressure is 7.1 atm

8 0
3 years ago
Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the h
andrew11 [14]

Answer : The change in entropy is 6.05\times 10^3J/K

Explanation :

Formula used :

\Delta S=\frac{m\times L_v}{T}

where,

\Delta S = change in entropy = ?

m = mass of water = 1.00 kg

L_v = heat of vaporization of water = 2256\times 10^3J/kg

T = temperature = 100^oC=273+100=373K

Now put all the given values in the above formula, we get:

\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}

\Delta S=6048.25J/K=6.05\times 10^3J/K

Therefore, the change in entropy is 6.05\times 10^3J/K

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MrRissso [65]
2 atoms of Al
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7 0
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