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svlad2 [7]
3 years ago
8

How may moles are in 88.4 grams of Al(OH)3?

Chemistry
1 answer:
mash [69]3 years ago
5 0

Answer:

1.133mol Al(OH)3

Explanation:

88.4/78.003 = 1.133

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ICE Princess25 [194]

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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