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3 years ago

On a potential energy diagram for the following processes, which of the following has an increase in entropy?

Chemistry

2 answers:

daser333 [38]3 years ago

7 0

Hello,

I believe that your answer would be B. water freezes
Hope this helps

Vaselesa [24]3 years ago

3 0

Answer: A. ice melts

Explanation: Entropy is the measure of randomness. A molecule with high disorder has high entropy.

When ice melts, solid changes to liquid and thus the molecules become more disordered.

When water freezes, liquid changes to solid, and thus the molecules become more ordered.

When steam condenses, gas changes to liquid and thus molecules become more ordered.

When water cools down in a refrigerator, liquid changes to solid, and thus the molecules become more ordered.

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tia_tia [17]

Heat can be absorbed or produced

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3 years ago

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2 years ago

Help! ASAP! Please on the questions

KiRa [710]

Answer:

62.15 m (squared)

Explanation:

We know that the Area of the square is 5.5 x 4= 22

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8 0

3 years ago

Read 2 more answers

Why can a solution be classified as a mixture

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Because they’re both made up of two substances that are not chemically combined

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3 years ago

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In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .

5 0

2 years ago