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ki77a [65]
3 years ago
5

What is the value of for this aqueous reaction at 298 K? A+B↽⇀C+D ΔG°=12.86 kJ/mol K=

Chemistry
1 answer:
kogti [31]3 years ago
5 0

Answer:

Kc = 0.5951 (4 sig. figs.)

Explanation:

For A + B ⇄ C + D at standard thermodynamic conditions (298K, 1atm)

ΔG = ΔG° + R·T·lnQ => 0 = ΔG° + R·T·lnKc => ΔG° = - R·T·lnKc

=> lnKc = - ΔG°/R·T

ΔG° = +12.86 Kj/mol

R = 8.314 Kj/mol·K

T = 298K

lnKc = - (+12.86Kj) / (8.314Kj/mol·K)(298K) = - 0.519 mol⁻¹

Kc = e⁻⁰°⁵¹⁹ mol⁻¹ = 0.5957 mol⁻¹ (4 sig. figs.)

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physical change because the gaseous water is chemically the same as the liquid

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A physical change can be defined as a type of change that only affects the physical form of a chemical substance (matter) without having any effect on its chemical properties. Thus, a physical change would only affect the physical appearance and properties of a chemical substance (matter) but not its chemical properties.

This ultimately implies that, a physical change result in a change of matter from one form or phase (liquid, solid or gas) to another without a corresponding change in chemical composition.

Hence, the boiling of water is considered to be a physical change because the gaseous water is chemically the same as the liquid i.e there isn't any changes in chemical composition of water when boiling.

8 0
2 years ago
Last year Steve was shorter than his brother. This year Steve is taller than his brother. What cell process is directly responsi
lutik1710 [3]

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5 0
3 years ago
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3 years ago
Can cells that are haploid (single set of chromosomes ) undergo meiosis
Dima020 [189]

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Hope this helps.

5 0
3 years ago
Need help !!!!! ASAP
gavmur [86]
<h2>Hello!</h2>

The answer is:

The new temperature will be equal to 4 K.

T_{2}=4K

<h2>Why?</h2>

We are given the volume, the first temperature and the new volume after the gas is compressed. To calculate the new temperature after the gas was compressed, we need to use Charles's Law.

Charles's Law establishes a relationship between the volume and the temperature at a gas while its pressure is constant.

Now, to calculate the new temperature we need to assume that the pressure is kept constant, otherwise, the problem would not have a solution.

From Charle's Law, we have:

\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

So, we are given the following information:

V_{1}=500mL\\T_{1}=20K\\V_{2}=100mL

Then, isolating the new temperature and substituting the given information, we have:

\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

T_{2}=\frac{T_{1}}{V_{1}}*V_{2} \\

T_{2}=\frac{20.00K}{500mL}*100mL\\

T_{2}=4K

Hence, the new temperature will be equal to 4 K.

T_{2}=4K

Have a nice day!

7 0
3 years ago
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