CHEMISTRY HIGH SCHOOL

Which phase change absorbs energy? A. H2O (g) --> H2O (s) B. H2O (g) --> H2O (l) C. H2O (l) --> H2O (g) D. H2O (l) --> H2O (s)

Answers

Answer 1
Answer:

Answer:

C. H2O (l) --> H2O (g)

Explanation:

Hello,

In this case, we need to remember that processes absorbing energy are those that have positive heat, it means, process that obtain energy from a certain source. In such a way, processes that need energy, in terms of phase chance are fusion, sublimation and evaporation, which are the change from solid to liquid, solid to gas and liquid to gas respectively. Therefore, for the given options, we can see that C. H2O (l) --> H2O (g) accounts for an evaporation process which actually absorbs energy.

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What determines the strength of a dipole-dipole force?A. The more symmetrical the molecule, the stronger the force.
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Answers

The more polar the molecules, the stronger the force. Hence, option C is correct.

What is dipole-dipole force?

Dipole-Dipole forces are the interaction between molecules of the permanent dipole. It occurs between the partially charged positive molecules and partially charged negative molecules.

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Learn more about dipole-dipole force here:

brainly.com/question/14195217

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The answer is C: the more polar the molecules, the stronger the force. Hope this helps!

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Answers

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Answers

Answer:

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Explanation:

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For each of the following strong base solutions, determine [OH−],[H3O+], pH, and pOH of 1.13×10^−2 M Ba(OH)2, 1.8x10^-4 M KOH, and 5.3x10^-4 M Ca(OH)2

Answers

pH calculations of strong base solutions are pretty direct. It all depends on how many OH- ions are dissociated and the concentration of the solution.

Ba(OH)2:

Since there are 2 OH- ions, the concentration of [OH-] is twice the concentration of the solution. Take the pOH by using the equation pOH = -log[OH-]. From here, you can get the pH by subtracting the pOH from 14. Finally, calculate the [H3O+] or [H+] concentration by using the equation pH = -log[H+] or [H+] = 10^(-pH)

[OH-] = 2 x 1.13x10^-2 M = 2.26x10^-2
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Follow the same rule for the other compounds.

KOH:

[OH-] = 1.8x10^-4 M
pOH = -log[OH-] = 3.74
pH = 14 - 3.74 = 10.26
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Ca(OH)2:
[OH-] = 2 x 5.3x10^-4 M = 1.06x10^-3 M
pOH = -log[OH-] = 2.97
pH = 14 - 2.97 = 11.03
[H3O+] = 10^(-11.03) = 9.33x10^-12
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