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3 years ago

How do you solve this ??

Chemistry

1 answer:

ICE Princess25 [194]3 years ago

8 0

Answer:

Option D. 400 mmHg

Explanation:

The following data were obtained from the question:

Mole of He (nHe) = 0.04 mole

Mole of Ne (nNe) = 0.06 mole

Total pressure = 10³ mmHg

Partial pressure of He =.?

Next, we shall determine the total number of mole in the reaction vessel.

This can be obtained as follow:

Mole of He (nHe) = 0.04 mole

Mole of Ne (nNe) = 0.06 mole

Total mole =?

Total mole = nHe + nNe

Total mole = 0.04 + 0.06

Total mole = 0.1

Next, we shall determine the mole fraction of He.

This can be obtained as follow:

Mole fraction = mole of gas /total mole

Mole of He (nHe) = 0.04 mole

Total mole = 0.1

Mole fraction of He =.?

Mole fraction of He = nHe/total mole

Mole fraction of He = 0.04/0.1

Mole fraction of He = 0.4

Finally, we shall determine the partial pressure of He as follow:

Partial pressure = mole fraction x total pressure

Mole fraction of He = 0.4

Total pressure = 10³ mmHg

Partial pressure of He =.?

Partial pressure of He = 0.4 x 10³

Partial pressure of He = 400 mmHg.

Therefore, the partial pressure of He is 400 mmHg.

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2 years ago

How many grams of KCl (s) has been produced from the thermal decomposition of KClO3 (s) that produced 50.0 mL of O2 (g) at 25 de

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Answer:

The mass is $m_{KCl} = 0.102 \ g$

Explanation:

From the question we are told that

The volume of oxygen produced is $V_o = 50mL = 50 *10^{-3} L$

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From the ideal gas law we have that

$PV = nRT$

Where R is the gas constant with the value

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n is the number of moles making it the subject of the formula

$n = \frac{PV}{RT}$

Substituting values

$n = \frac{1 * (50 *10^{-3}) }{0.08206 * 298}$

$n = 2.045 *10^{-3} \ mol$

From the chemical equation

one mole of $KClO_3$ produces one mole of kCl and $\frac{3}{2}$ of oxygen

x mole of $KClO_3$ produces x mole of kCl and $2.045 *10^{-3} \ mol$ of oxygen

So $x = \frac{2.045 *10^{-3}}{\frac{3}{2} }$

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Now the molar mass of KCl is a constant with a value

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$m_{KCl} = x * M_{KCl}$

Substituting values

$m_{KCl} = 1.363 *10^{-3} * 74.55$

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3 years ago