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Ann [662]
3 years ago
8

A container of a mixture of 4 gases has a total pressure of 35.7 kPa. Gas A has a partial pressure of 7.8kPa. Gas B has a partia

l pressure of 3.7 kPa, and Gas C has a pressure of 8.7kPa. What is the partial pressure of Gas D in kPa?
Chemistry
1 answer:
Maurinko [17]3 years ago
4 0

Answer:

partial pressure of gas D Pd = 15.5 kPa

Explanation:

As per the Dalton's law of partial pressure, in a mixture, pressure exerted by each gas when summed gives the total partial pressure exerted by mixture.

P(Total) = P1+P2+P3.....

Given P(Total) = 35.7 kPa

Partial pressure of gas A Pa = 7.8 kPa

Partial pressure of gas B Pb = 3.7 kPa

Partial pressure of gas C Pc =  8.7 kPa

There, Partial pressure of gas D Pd = P(Total) -(Pa+Pb+Pc)

Pd = 35.7-(7.8+3.7+8.7) = 35.7-20.2 kPa = 15.5 kPa

Therefore, partial pressure of gas D Pd = 15.5 kPa

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The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio
iren2701 [21]

Answer : The correct option is, (B) CO_2

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of unknown gas = 11.9\text{ mL }min^{-1}

R_2 = rate of effusion of oxygen gas = 14.0\text{ mL }min^{-1}

M_1 = molar mass of unknown gas  = ?

M_2 = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}

M_1=44.2g/mole

The unknown gas could be carbon dioxide (CO_2) that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide (CO_2)

5 0
3 years ago
Which is the best location for storing radioactive wastes?
vesna_86 [32]
<span>Deep geological disposal is widely agreed to be the best solution for final disposal of the most radioactive waste produced.

</span>Geological disposal<span> involves isolating radioactive waste </span>deep<span> inside a suitable rock volume to ensure that no harmful quantities of radioactivity ever reach the surface environment.
</span>
Hope this helps :)
4 0
3 years ago
Read 2 more answers
[02.01]Which of the following is essential for a theory to become widely accepted within the scientific community?
mezya [45]
I think maybe empirical evidence
4 0
3 years ago
Pyridinium is a weak acid having a pKa of 5.2. How much pyridine (the conjugate base of pyridinium) must be added to an aqueous
Whitepunk [10]

Answer:

Amount of pyridine required = 0.0316 M

Explanation:

pH of a buffer solution is calculated by using Henderson - Hasselbalch equation.

pH=pK_a+log\frac{[Conjugate\ base]}{[weak\ acid]}

Pyridinium is a weak acid and in the presence of its conjugate base, it acts as buffer.

Henderson - Hasselbalch equation for pyridine/pyridinium buffer is as follows:

pH=pK_a+log\frac{[Py]}{PyH^+]}

pH = 4.7

pK_a=5.2

PyH^+ (Pyridinium)=0.100 M

Substitute the values in the formula

pH=pK_a+log\frac{[Py]}{PyH^+]}\\4.7=5.2 log\frac{[Py]}{0.100}

4.7-5.2=log\frac{[Py]}{0.100} \\-0.5=log\frac{[Py]}{0.100}\\\frac{[Py]}{0.100}=antilog -0.5\\\frac{[Py]}{0.100}=0.316

\frac{[Py]}{0.100} =0.316

[Py]=0.0316\ M

Amount of pyridine required = 0.0316 M

7 0
3 years ago
What volume would 7.83 X 10^3 moles of Nitrogen gas occupy?
Kay [80]

Answer:

29.232 l

Explanation:

6.023x10^23 mole of nitrogen =22.4 l

7.83×10^23 mole of nitrogen =

22.4/(6.023×10^23)×7.83×10^23

=29.232 l

4 0
3 years ago
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