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olga nikolaevna [1]
2 years ago
13

A gas has a volume of 22.4 L at 0.853 atm. What pressure is needed to change the volume to 24.0L?

Chemistry
1 answer:
Scrat [10]2 years ago
6 0

The pressure needed to change the volume to 24 L is 0.796 atm.

Explanation:

It is known by Boyle's law that the pressure experienced by the gas molecules will be inversely proportional to the volume occupied by the molecules.

P =\frac{1}{V}

So as the initial volume is said to be 22.4 L, consider it as V₁ = 22.4 L. Then the initial pressure is said to be 0.853 atm, so P₁ = 0.853 atm. So we have to determine the new pressure P₂ when the volume is changed to V₂ = 24 L.  As there is increase in the volume, the pressure should be decreased due to Boyle's law. Thus, as per Boyle's law, the two pressures and their volumes can be related as

P_{1} V_{1} =  P_{2} V_{2}

0.853*22.4 = P_{2}*24\\\\P_{2} = \frac{0.853*22.4}{24} = 0.796 atm

Thus, the pressure gets decreased to 0.796 atm on increase in the volume to 24 L.

So the pressure needed to change the volume to 24 L is 0.796 atm.

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A 220.0 gram piece of copper is dropped into 500.0 grams of water 24.00 °C. If the final temperature of water is 42.00 °C, what
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Answer:

C. 481 °C.

Explanation:

  • At equilibrium:

The amount of heat absorbed by water = the amount of heat released by copper.

  • To find the amount of heat, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of energy.

m is the mass of substance.

c is the specific heat capacity.

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T).

<em>∵ Q of copper = Q of water</em>

∴ - (m.c.ΔT) of copper = (m.c.ΔT) of water

m of copper = 220.0 g, c of copper = 0.39 J/g °C, ΔT of copper = final T - initial T = 42.00 °C - initial T.

m of water = 500.0 g, c of water = 4.18 J/g °C, ΔT of water = final T - initial T = 42.00 °C - 24.00 °C = 18.00 °C.

∴ - (220.0 g)( 0.39 J/g °C)(42.00 °C - Ti) = (500.0 g)(4.18 J/g °C)(18.00 °C)

∴ - (85.8)(42.00 °C - Ti) = 37620.

∴ (42.00 °C - Ti) = 37620/(- 85.8) = - 438.5.

∴ Ti = 42.00 °C + 438.5 = 480.5°C ≅ 481°C.

<em>So, the right choice is: C. 481 °C.</em>

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Explanation:

Acidity of amine derivatives can derived from their pKa values.

The rule of thumb for acidity with relation to pKa values is that:

As the pKa decreases the acid strength increases and the conjugate base decreases. Similarly, as the pKa increases, the acid strength decreases and the conjugate base increase.

Hence the stronger the acid , the  lower pKa value  and the weaker the acid , the stronger the pKa value.

So the pKa value for anilinium ion = 4.6

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Similarly, for aniline and secondary amine, in order to determine the derivative with the higher acidity, we will consider the electron withdrawing substituent group.

The more difficult the electron are being withdraw from the electron withdrawing substituent , the more acidic the compound.

In aniline , the stabilized benzene ring attached to NH₂ makes it a less electron withdrawing group compared to the straight chains structure found in secondary amine where electron are easily withdraw by nucleophilic substitution reactions.

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Answer:

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Explanation:

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