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Nataly [62]
3 years ago
14

A gas has a pressure of 1.00 atm and a volume of 500 mL. When its pressure is doubled, at constant temperature, to 2.00 atm, its

volume halves to 250 mL, following Boyle's Law. However, when its pressure is increased 1000 times, at constant temperature, to 1000 atm, the volume decreases to 2 mL which is more than 1/1000 of 500 mL. What is the reason why Boyle's Law is not obeyed when the pressure is increased to 1000 atm?
Chemistry
1 answer:
Advocard [28]3 years ago
3 0

Answer: The reason why Boyle's is not obeyed is because the size of the gaseous molecules is more appreciable compared to the volume of the container containing the gas.

Explanation: At such high pressure of 1000atm , there will be high amount of gaseous molecules in terms of numbers and the volume of the gas will not be able to be decreased as if the gaseous molecules were dimensionless. The piston of the container containing the gas will not be able to pushed down much because the molecules will require an appreciable space.

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Answer:

41.64mL of NaOH 0.500M must be added to obtain the desire pH

Explanation:

It is possible to find pH of a buffer by using H-H equation, thus:

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<em>Where [HA] is concentration of the weak acid TRIS-HCl and [A⁻] is concentration of its conjugate acid.</em>

Replacing in H-H equation:

7.60 = 8.072 + log [A⁻] / [HA]

0.3373 =  [A⁻] / [HA] <em>(1)</em>

10.0g of TRIS-HCl (Molar mass: 121.135g/mol) are:

10.0g ₓ (1mol / 121.135g) = 0.08255 moles of acid. That means moles of both the acid and conjugate base are:

[A⁻] + [HA] = 0.08255 <em>(2)</em>

Replacing (1) in (2):

0.3373 =  0.08255 - [HA] / [HA]

0.3373[HA] =  0.08255 - [HA]

1.3373[HA] = 0.08255

<em>[HA] = 0.06173 moles</em>

Thus:

[A⁻]  = 0.08255 - 0.06173 = 0.02082 moles [A⁻]

The moles of A⁻ comes from the reaction of the weak acid with NaOH, that is:

HA + NaOH → A⁻ + H₂O + K⁺

Thus, <em>you need to add 0.02082 moles of NaOH to produce 0.02082 moles of A⁻. </em>As NaOH solution is 0.500M:

0.02082 moles NaOH ₓ (1L / 0.500mol) = 0.04164L of NaOH 0.500M =

<h3>41.64mL of NaOH 0.500M must be added to obtain the desire pH</h3>

3 0
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Answer:

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Answer:

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