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Ganezh [65]
3 years ago
15

A bottle of 2 moles of an ideal diatomic gas experiences a temperature increase of 50.0K at constant volume. (a) Find the increa

se in internal energy if only translational and rotational motions are possible. (b) Find the increase in internal energy if translational, rotational, and vibrational motions are possible. (c) Which of the following kind of molecule is more difficult to increase temperature with the same amount of heat: (i) molecule that has many vibrational and rotational modes or (ii) molecule with only one of such modes?
Chemistry
1 answer:
andre [41]3 years ago
5 0

Answer:

Explanation:

a) For diatomic gas: Translational motion = 3 and rotational motion = 2

∴ Total (internal energy) = 3 + 2 = 5

b) Translational + Rotational + Vibrational = 3 + 2 + 1 = 6

c) Linear molecule  

i) Non linear molecule

ii)  Monatomic molecule

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What is the ph value of the produced solution when hcl reacts with naoh ?
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You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution when ex
mojhsa [17]

Answer:

a. 1.78x10⁻³ = Ka

2.75 = pKa

b. It is irrelevant.

Explanation:

a. The neutralization of a weak acid, HA, with a base can help to find Ka of the acid.

Equilibrium is:

HA ⇄ H⁺ + A⁻

And Ka is defined as:

Ka = [H⁺] [A⁻] / [HA]

The HA reacts with the base, XOH, thus:

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As you require 26.0mL of the base to consume all HA, if you add 13mL, the moles of HA will be the half of the initial moles and, the other half, will be A⁻

That means:

[HA] = [A⁻]

It is possible to obtain pKa from H-H equation (Equation used to find pH of a buffer), thus:

pH = pKa + log₁₀ [A⁻] / [HA]

Replacing:

2.75 = pKa + log₁₀ [A⁻] / [HA]

As [HA] = [A⁻]

2.75 = pKa + log₁₀ 1

<h3>2.75 = pKa</h3>

Knowing pKa = -log Ka

2.75 = -log Ka

10^-2.75 = Ka

<h3>1.78x10⁻³ = Ka</h3>

b. As you can see, the initial concentration of the acid was not necessary. The only thing you must know is that in the half of the titration, [HA] = [A⁻]. Thus, the initial concentration of the acid doesn't affect the initial calculation.

7 0
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