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qwelly [4]
3 years ago
8

Why is it difficult to classify dissolving as simply a physical or a chemical change? (1 point)

Chemistry
2 answers:
Likurg_2 [28]3 years ago
6 0

The correct answer is:Some substances keep the same molecular structure when they break down, and others do not.

Some substances change their form when they dissolve and some do not. Dissolution can sometimes be regarded as a sort of reaction between a chemical substance and water.

Usually, we can consider dissolution of a substance in water as a sort of chemical reaction for some substances. For instance, an ionic substance interacts with water to form ions. similarly, some salts become hydrolysed in water and give acidic/basic solutions as  result of that.

However, some substances do not interact with water upon dissolution. They rather remain as molecular entities because they are not composed of ions.

We can see that some substances keep the same molecular structure when they break down, and others do not keep the same molecular structure  when they dissolve hence it is difficult to classify dissolving as a physical or a chemical change.

brainly.com/question/1161517

MariettaO [177]3 years ago
6 0

Answer:

It is difficult because of C

Explanation:

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What is the reduction potential of a hydrogen electrode that is still at standard pressure, but has ph = 5.65 , relative to the
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Answer:

\boxed{\text{-0.275 V}}

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1. Write the equation for the cell reaction

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T = 25 + 273.15 = 298.15 K

\text{H}^{+} = 10^{\text{-pH}} = 2.24 \times 10^{-5}\text{ mol/\L}\\\\Q = \dfrac{\text{[H}^{+}]_{\text{prod}}^{2}}{\text{[H}^{+}]_{\text{react}}^{2}} = \dfrac{(1.00)^{2}}{(2.24 \times 10^{-5})^{2}} =2.00 \times 10^{9}\\\\\\E = 0 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln{2.00 \times 10^{9}}\\\\= -0.01285 \times 21.41 = \textbf{-0.275 V}\\\text{The cell potential for the cell as written is }\boxed{\textbf{-0.275 V}}

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