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jeka94
2 years ago
7

By what factor would the average kinetic energy of the particles by increased if the temperature of a substance was increased fr

om -49°C to 287°C?
Chemistry
1 answer:
natita [175]2 years ago
3 0
Average kinetic energy of a particle :
0.5 mv^2 = kT/2
so the kinetic energy = kT/2
assuming the same value of K
T1 = -49 + 273 = 224
T2 = 287 + 273 = 560

E2 / E1 = kT2 / 2 / kT1 / 2
E2 / E1 = T2 / T1
E2 / E1 = 560 / 224 = 2.5
so the average kinetic energy of the particle increases by 2.5
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What is the expected value for the heat of sublimation of acetic acid if its heat of fusion is 10.8 kJ/mol and its heat of vapor
Dennis_Churaev [7]

Answer:

35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

Explanation:

CH_3COOH(l)\rightarrow CH_3COOH(g)..[1]

Heat of vaporization of acetic acid = H^o_{vap}=24.3 kJ/mol

CH_3COOH(s)\rightarrow CH_3COOH(l)..[2]

Heat of fusion of acetic acid = H^o_{fus}=10.8 kJ/mol

Heat of sublimation of acetic acid = H^o_{sub}=?

CH_3COOH(s)\rightarrow CH_3COOH(g)..[3]

[1] + [2] = [3] (Hess's law)

H^o_{sub}=H^o_{vap}+H^o_{fus}

=24.3 kJ/mol+10.8 kJ/mol=35.1 kJ/mol

35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

5 0
2 years ago
A mixture with H2 and He exerts a total pressure of 0.48 atm. If there is 1.0 g of H2 and 1.0 g of He in the mixture, what is th
lara [203]

Answer is: the partial pressure of the helium gas is 0.158 atm.

p(mixture) = 0.48 atm; total pressure.

m(H₂) = 1.0 g; mass of hydrogen gas.

n(H₂) = m(H₂) ÷ M(H₂).

n(H₂) = 1.0 g ÷ 2 g/mol.

n(H₂) = 0.5 mol; amount of hydrogen.

m(He) = 1.0 g; mass of helium.

n(He) = 1 g ÷ 4 g/mol.

n(He) = 0.25 mol; amount of helium.

χ(H₂) = 0.5 mol ÷ 0.75 mol.

χ(H₂) = 0.67; mole fraction of hydrogen.

χ(He) = 0.25 mol ÷ 0.75 mol.

χ(He) = 0.33; mole fraction of helium.

p(He) = 0.33 · 0.48 atm.

p(He) = 0.158 atm; the partial pressure of the helium gas.

8 0
3 years ago
The magnetic field of an object exerts a force on any magnetic object around it.
nika2105 [10]
I believe the answer is true
3 0
2 years ago
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If 20.0 mL of glacial acetic acid (pure HC2H3O2HC2H3O2) is diluted to 1.70 L with water, what is the pH of the resulting solutio
Andreas93 [3]

Answer:

The answer to your question is pH = 0.686

Explanation:

Data

Acetic acid = 20 ml

Final volume = 1.7 L

pH = ?

density = 1.05 g/ml

Process

1.- Calculate the mass of acetic acid

density = mass / volume

mass = density x volume

mass = 1.05 x 20

mass = 21 g

2.- Calculate the moles of acetic acid (CH₃COOH)

Molar mass = (12 x 2) + (16 x 2) + (4 x 1)

                   = 24 + 32 + 4

                   = 60 g

                 60 g of acetic acid ---------------- 1 mol

                 21 g of acetic acid ----------------- x

                 x = (21 x 1) / 60

                 x = 0.35 moles

3.- Calculate the concentration

Molarity = 0.35 moles / 1.70 L

              = 0.206

4.- Calculate the pH

pH = -log[0.206]

pH = 0.686

6 0
3 years ago
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