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3 years ago

What is the change in internal energy ( ΔU ) of the system if q = –8 kJ and w = –1 kJ for a certain process?

Chemistry

1 answer:

Radda [10]3 years ago

6 0

Answer:

Change in internal energy (ΔU) = -9 KJ

Explanation:

Given:

q = –8 kJ [Heat removed]

w = –1 kJ [Work done]

Find:

Change in internal energy (ΔU)

Computation:

Change in internal energy (ΔU) = q + w

Change in internal energy (ΔU) = -8 KJ + (-1 KJ)

Change in internal energy (ΔU) = -8 KJ - 1 KJ

Change in internal energy (ΔU) = -9 KJ

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Part b an "empty" container is not really empty if it contains air. how may moles of nitrogen are in an "empty" two-liter cola b

Sedbober [7]

6.38x10^-2 moles
First, let's determine how many moles of gas particles are in the two-liter container. The molar volume for 1 mole at 25C and 1 atmosphere is 24.465 liters/mole. So
2 L / 24.465 L/mol = 0.081749438 mol
Now air doesn't just consist of nitrogen. It also has oxygen, carbon dioxide, argon, water vapor, etc. and the total number of moles includes all of those other gasses. So let's multiply by the percentage of nitrogen in the atmosphere which is 78%
0.081749438 mol * 0.78 = 0.063764562 mol.
Rounding to 3 significant figures gives 6.38x10^-2 moles

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2 years ago

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MatroZZZ [7]

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2 years ago

A double covalent bond involves two atoms sharing two electrons? True or false

jarptica [38.1K]

Answer:

False

Explanation:

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*single covalent bond shares 2 electrons.

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3 years ago

Which phase of matter does not change shape or volume? *

RideAnS [48]

Answer:

Solid is ur answer

Explanation:

Stay safe, stay healthy and blessed.

Have a great day !

Thank you

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2 years ago

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Assume that silver and gold form ideal, random mixtures. Calculate the mass of pure Ag needed to cause an entropy increase of 20

KengaRu [80]

Answer:

$m_{Ag}=2,265.9g$

Explanation:

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In this case, since the definition of entropy in a random mixture is:

$\Delta S=-n_TR\Sigma[x_i*ln(x_i)]$

For this silver-gold mixture we write:

$\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]$