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Allushta [10]
2 years ago
14

Calculate q, the heat released in each reaction.

Chemistry
1 answer:
goldenfox [79]2 years ago
8 0

First Set of Answers:

Heat Released Reaction 1: 3700

Heat Released Reaction 2: 3200

Second Set of Answers:

Moles Reactant 1: .00823

Moles Reactant 2: .0397

Third Set of Answers:

Enthalpy 1: -450

Enthalpy 2: -81

FOR EDG!!!!!!!!!!!!

( Record in your data table (; )

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Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar
kaheart [24]

This question is incomplete, the complete question is;

Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar, 520 K. If the carbon dioxide behaves as an ideal gas, determine the amount of entropy produced, in kJ/K. Assuming;

a) constant specific heats Cp = 0.939 kJ/Kg K

b) variable specific heats

Answer:

a) the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced is 0.69845 kJ/K

Explanation:

Given the data in the question;

5 kg of carbon dioxide (CO₂) gas undergoes a process in a well-insulated piston-cylinder assembly.

m = 5 kg

Molar mass M = 44.01 g/mol

P₁ = 2 bar, P₂ = 20

T₁ = 280 K, P₂ = 520 K

Since its insulated { q = 0 } ( kinetic and potential energy effects = 0 )

Now,

a) the amount of entropy produced, in kJ/K, Assuming constant specific heats with Cp = 0.939 kJ/Kg K

S_{Generation = m × ((Cp × In( T₂/T₁) - R × In( P₂/p₁ ))

we substitute

S_{Generation = 5 × (( 0.939  × In( 520/280) - 0.1889 × In( 20/2 ))

= 5 × ( 0.5812778 - 0.434958 )

= 5 × 0.1463198

= 0.731599 kJ/K

Therefore, the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced, in kJ/K, Assuming variable specific heats.

Now, from  Table A-23: Ideal Gas Properties of Selected Gases;

T₁,T₂ : s₁⁰ = 211.376 kJ/kmol-K, s₂⁰ = 236.575 kJ/kmol-K

now, s₁ = s₁⁰ / M and s₂ = s₂⁰ / M

we substitute

s₁ = s₁⁰ / M = 211.376 / 44.01  = 4.8029 kJ/kg

s₂ = s₂⁰ / M = 236.575 / 44.01 = 5.37548 kJ/kg

S_{Generation = m × (( s₂ - s₁ ) - R × In( p₂ / p₁ ))

we substitute

S_{Generation = 5 × (( 5.37548 - 4.8029  ) - 0.1880 × In( 20 / 2 ))

= 5 × ( 0.57258 - 0.432885997 )

= 5 × 0.13969

= 0.69845 kJ/K

Therefore, the amount of entropy produced is 0.69845 kJ/K

5 0
2 years ago
How many milliliters of a 0.266 m lino3 solution are required to make 150.0 ml of 0.075 m lino3 solution?
Allisa [31]

We need an equation that would relate the concentration of the original solution to that of the desired solution. To solve this we use the equation expressed as follows, 

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

M1V1 = M2V2

0.266 M x V1 = 0.075 M x 150 mL

V1 = 42.29 mL


Therefore, we need about 42.29 mL of the 0.266 M of lithium nitrate solution to make 150.0 mL of the 0.075 M lithium nitrate solution.

3 0
3 years ago
Read 2 more answers
Classify the following materials as solid, liquid, or gas at room temperature: milk, helium, granite, oxygen, steel, and gasolin
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Im not sure but milk is liquid, helium is gas, granite is solid, oxygen is gas, steel is solid, and gasiloine is liquid.
3 0
3 years ago
Does a Chemical reaction happen between Sodium and water? I NEED the answer NOW! (please!!!)
kompoz [17]
A chemical reaction does happen between sodium and water to form Sodium Hydroxide (NaOH) and Hydrogen Gas (H-2)
5 0
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Consider two gases, A and B, are in a container at room temperature. What effect will the following changes have on the rate of
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Answer:

  1. decrease in temperature , decreases the kinetic movements of the gase molecules as a result decreases the frequency of collisions between gas molecules A and B consequently decreases the rate of reactions of gases A and B
6 0
2 years ago
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