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PSYCHO15rus [73]
3 years ago
9

What is the final volume of a sample of neon under 5.6 atm of pressure at 250 k if it was initially at 6.7 atm or pressure, 500k

, and was 18L?
Chemistry
2 answers:
icang [17]3 years ago
3 0

Answer:

V_2=10.8L

Explanation:

Hello,

In this case, we use the combined idea gas equation that help us to understand the volume-pressure-temperature relationship as:

\frac{P_1V_1}{T_1}= \frac{P_2V_2}{T_2}

So, we solve for the final volume as shown below, considering the given data at the beginning of the experiment:

V_2=\frac{P_1V_1T_2}{T_1P_2}=\frac{6.7atm*18L*250K}{500K*5.6atm}\\ \\V_2=10.8L

Best regards.

katrin [286]3 years ago
3 0

Answer:

10.77 L

Explanation:

Initial pressure P1= 6.7atm

Initial temperature T1= 500K

Initial volume V1= 18L

Final pressure P2= 5.6 atm

Final temperature T2= 250K

Final volume V2=???

From the general gas equation:

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2= 6.7×18×250/5.6×500= 30150/2800

V2= 10.77 L

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Use standard enthalpies of formation to calculate Δ H ∘ rxn for each reaction. MISSED THIS? Read Section 7.9; Watch KCV 7.9, IWE
Eva8 [605]

Answer:

Standard Heat of Reaction 1 = -136.2 kJ/mol

Standard Heat of Reaction 2 = -41.166 kJ/mol

Standard Heat of Reaction 3 = -136.07 kJ/mol

Standard Heat of Reaction 4 = 279.448kJ/mol

Explanation:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

The required standard heat of formation for each of the reactants and product above, as obtained from literature is listed below.

C₂H₄ (g), 52.5 kJ/mol

H₂ (g), 0 kJ/mol

C₂H₆ (g), -83.7 kJ/mol

CO (g), -110.525 kJ/mol

H₂O (g), -241.818 kJ/mol

H₂ (g), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

NO₂ (g), 33.2 kJ/mol

H₂O (l), -285.8 kJ/mol

HNO₃ (aq), -206.28 kJ/mol

NO (g), 90.29 kJ/mol

Cr₂O₃ (s), -1128.4 kJ/mol

CO (g), -110.525 kJ/mol

Cr (s), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

Note that

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×-83.7) = -83.7 kJ/mol

ΔH∘(reactants) = (1×52.5) + (1×0) = 52.5 kJ/mol

ΔH∘(rxn) = -83.7 - 52.5 = -136.2 kJ/mol

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×0) + (1×-393.509) = -393.509 kJ/mol

ΔH∘(reactants) = (1×-110.525) + (1×-241.818) = -352.343 kJ/mol

ΔH∘(rxn) = -393.509 - (-352.343) = -41.166 kJ/mol

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×-206.28) + (1×90.29) = -322.27 kJ/mol

ΔH∘(reactants) = (3×33.2) + (1×-285.8) = -186.2 kJ/mol

ΔH∘(rxn) = -322.27 - (-186.2) = -136.07 kJ/mol

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×0) + (3×-393.509) = -1,180.527 kJ/mol

ΔH∘(reactants) = (1×-1128.4) + (3×-110.525) = -1,459.975 kJ/mol

ΔH∘(rxn) = -1,180.527 - (-1,459.975) = 279.448 kJ/mol

Hope this Helps!!!

4 0
3 years ago
A scientist has 0.12 miles of a gas at a pressure of 4.06 kPa and a volume of 14 liters in an enclosed container. What is the te
natka813 [3]

Answer:

56972.17K

Explanation:

P = 4.06kPa = 4.06×10³Pa

V = 14L

n = 0.12 moles

R = 8.314J/Mol.K

T = ?

We need ideal gas equation to solve this question

From ideal gas equation,

PV = nRT

P = pressure of the ideal gas

V = volume the gas occupies

n = number of moles

R = ideal gas constant

T = temperature of the gas

PV = nRT

T = PV / nR

T = (4.06×10³ × 14) / (0.12 × 8.314)

T = 56840 / 0.99768

T = 56972.17K

Note : we have a large number for temperature because we converted the value of pressure from kPa to Pa

7 0
3 years ago
Which best describes the transition from gas to liquid?
Nina [5.8K]

A - its condensation and gas particles have a higher kinetic energy

8 0
3 years ago
Read 2 more answers
What is similar about the outer shell of electrons in the alkali metals family
Nadusha1986 [10]

Answer:

1 electron

Explanation:

These metals have a single electron in the outer shell

4 0
3 years ago
The electron configuration for an element in the halogen group should always end with Question 16 options:
ludmilkaskok [199]

Answer:

ns²np⁵

Explanation:

The electronic configuration of elements of halogen group ends with 7 electrons in its outer most shell and they distributed as ns²np⁵.

Such as Cl (17 electron): 2,8,7 which is 1s²2s²2p⁶3s²3p⁵.

5 0
3 years ago
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