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3 years ago

Calculate ∆E for the following case:

Chemistry

1 answer:

NARA [144]3 years ago

5 0

Answer:

The internal energy(ΔE) of a substance is calculated below:

From the first law of thermodynamics;

ΔE=q+w

Explanation:

ΔE=q+w

here;

+q=endothermic reaction

-q=exothermic reaction

+w=work done on the system

-w=work done by the system

Given:

q=+1.62kJ=1620J

w=-874J

To solve:

the internal energy(ΔE)

We know:

ΔE=q+w

according to the problem;

ΔE=q-w

since;

w=-874J (i.e.)work is done by the system.

ΔE=1620-874

ΔE=+746J

Therefore the internal energy is +746J

i.e. the option is "c"(+746J)

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A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 K. What is the pressure in atm calculated using t

soldi70 [24.7K]

Answer:

The pressure in atm calculated using the van der Waals' equation, is 337.2atm

Explanation:

This is the Van der Waals equation for real gases:

(P + a/v² ) ( v-b) = R .T

where P is pressure

v is Volume/mol

R is the gas constant and T, T° in K

a y b are constant for each gas, so those values are data, from the statement.

[P + 1.345 L²atm/mol² / (0.7564L/10.21mol)² ] (0.7564L/10.21mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

[P + 1.345 L²atm/mol² / 5.48X10⁻³ L²/mol²] (0.074 L/mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 24.34 L.atm/mol

0.04181L/mol .P + 10.24 L.atm/mol = 24.34 L.atm/mol

0.04181L/mol .P = 24.34 L.atm/mol - 10.24 L.atm/mol

0.04181L/mol. P = 14.1 L.atm/mol

P = 14.1 L.atm/mol / 0.04181 mol/L

P = 337.2 atm

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3 years ago

What is one molar volume of the gas ammonia (nh3) at stp?

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2 years ago

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ZanzabumX [31]

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3 years ago

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A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a

Nata [24]

Answer:

There are present 5,5668 moles of water per mole of CuSO₄.

Explanation:

The mass of CuSO₄ anhydrous is:

23,403g - 22,652g = 0,751g.

mass of crucible+lid+CuSO₄ - mass of crucible+lid

As molar mass of CuSO₄ is 159,609g/mol. The moles are:

0,751g × $\frac{1mol}{159,609g}$ = 4,7052x10⁻³ moles CuSO₄

Now, the mass of water present in the initial sample is:

23,875g - 0,751g - 22,652g = 0,472g.

mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid

As molar mass of H₂O is 18,02g/mol. The moles are:

0,472g × $\frac{1mol}{18,02g}$ = 2,6193x10⁻² moles H₂O

The ratio of moles H₂O:CuSO₄ is:

2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668

That means that you have 5,5668 moles of water per mole of CuSO₄.

I hope it helps!

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3 years ago