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2 years ago

Predict the sign of the entropy change, ΔS, for the following processes. Explain the reason for each of the predictions.

Chemistry

1 answer:

Nastasia [14]2 years ago

6 0

The sign of the entropy change, ΔS, for the following processes include:

I2 (s) + ½ Cl2 (g) ← → ICl (g) - positive

3 Ag (s) + 4 HNO3 (aq) → 3AgNO3 (aq) + NO (g) + 2 H2O (l) - positive

Cl2 (g) → Cl2 (l) - negative

C5H12 (g) + 8 O2 (g) → 5 CO2 (g) + 6 H2O (l) - negative

What is Entropy?

This is referred to the degree of randomness or disorderliness of a system and is denoted as S.

Entropy change is usually positive when solid or liquid reactant is converted into gas product and vice versa.

Read more about Entropy here brainly.com/question/419265

#SPJ1

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The volume of a gas with an initial pressure of 380 mmHg increases from 5.0 L to 8.0 L. What is the final pressure of the gas, i

Phoenix [80]

It’s C. 0.31 atm

I hope this helped out! Have a nice day :)

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3 years ago

Calculate the ph of the solution resulting by mixing 20.0 ml of 0.15 m hcl with 20.0 ml of 0.10 m koh

Aliun [14]

Answer:

1.60.

Explanation:

The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.

The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.

Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.

∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.

∵ pH = - log[H⁺]

∴ pH = - log[H⁺] = - log(0.025) = 1.602 ≅ 1.60.

5 0

3 years ago

What accounts for the large increase in volume when sugar burns?

Dimas [21]

I don’t know type it up yourself

8 0

3 years ago

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

4 0

3 years ago

what would be the value of 'G' on the surface of earth if it's mass was twice & its radius half of what it is how?

Katen [24]

Answer:

What would be the value of g on the surface of the earth if its mass was twice and its radius half of what it is now ? g2=8g1 Thus , the value of g on the surface of the earth would be eight times the present value.

Explanation:

Hope this helps! :)

8 0

3 years ago

Read 2 more answers