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umka2103 [35]
3 years ago
14

Determine the temperature of 2.49 moles of gas in a 1.0-L vessel at 143

Chemistry
1 answer:
bulgar [2K]3 years ago
5 0

Answer:

Temperature = 6.91K

Explanation:

Temperature = T

Number of Moles = 2.49moles

Volume = 1.0L

Pressure = 143kPa

R = 8.31kPa*L/mol.K

From ideal gas equation,

PV = nRT

P = pressure of the ideal gas

V = volume which the gas occupies

n = number of moles of the gas

R = ideal gas constant

T = temperature of the ideal gas

PV = nRT

T = PV / nR

T = (143 * 1) / (2.49 * 8.31)

T = 143 / 20.6919

T = 6.91K

The temperature of the gas is 6.91K

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Why does the melting of the polar ice caps facilitate the absorption of CO2 by ocean waters?​
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hi brainly user! ૮₍ ˃ ⤙ ˂ ₎ა

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Answer:

The melting of the caps increases the concentration of CO2 in the water, making its absorption faster.

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Explanation:

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The polar caps of our planet have in their composition several elements, among them the CO2 that is absorbed by the atmosphere. Cold waters, which are present in the Arctic, have an easier time absorbing CO2 compared to other waters.

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1 year ago
10) How many grams are there in 1.00 x 10^24 molecules of BC13?
lana66690 [7]

194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have \frac{1}{6.022*10^{23} } moles.

So 1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles

grams of BCl_{3}=117.17*1.66=194.5 g

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

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