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Sladkaya [172]
3 years ago
13

A chemist prepares a sample of helium gas at a certain pressure, temperature and volume and then removes all but a fourth of the

gas molecules (only a fourth remain). How must the temperature be changed (as a multiple of T1) to keep the pressure and the volume the same?
Chemistry
1 answer:
cupoosta [38]3 years ago
6 0

Answer:

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

Explanation:

Pressure in the container is P and volume is V.

Temperature of the helium gas molecules =T_1

Molecules helium gas = x

Moles of helium has = n_1= \frac{x}{N_A}

PV = nRT (Ideal gas equation)

PV=n_1RT_1...[1]  

After removal of helium gas only a fourth of the gas molecules remains and pressure in the container and volume should remain same.

Molecules of helium left after removal = \frac{x}{4}

Moles of helium has left after removal = n_2= \frac{x}{4\times N_A}

PV=n_2RT_2...[2]

n_1RT_1=n_2RT_2

\frac{x}{N_A}\times T_1=\frac{x}{4\times N_A}\times T_2

T_1=\frac{T_2}{4}

T_2=4T_1

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

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Answer:

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Explanation:

The weak acid with the lowest pKa will be the most acidic. In the other way, the conjugate base which the acid is weak will be strong.

The weak base with the lowest pKb will be the most basic. And the conjugate base of the weak base will be a strong acid.

Ka Acetic acid = 1.8x10-5

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NH4Br has the lowest pH

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The most basic between the conjugate base of the acetic acid, NaCH3CO2 and KCN is KCN because the acetic acid is the stronger acid regard to HCN.

The rank is:

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