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lina2011 [118]
3 years ago
6

Aluminum has a specific heat of 0.902 j/gC. How much heat is lost when a piece of aluminum with a mass of 23.984 g cools from a

temperature of 415 degrees to 22 degrees?
Chemistry
1 answer:
Marizza181 [45]3 years ago
8 0

Answer:

Q = - 8501.99 j

Explanation:

Given data:

Specific heat of Al = 0.902 j/g.°C

Heat lost = ?

Mass of sample = 23.984 g

Initial temperature = 415°C

Final temperature = 22°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 22°C - 415°C

ΔT = -393°C

Q = m.c. ΔT

Q = 23.984 g× 0.902 j/g.°C × -393°C

Q = - 8501.99 j

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Nucleus of tritium contains....neutrons......protons
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4 years ago
Differentiate between atoms, elements, molecules and compounds<br><br> *URGENT
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6 0
4 years ago
6. A sample of a gas at 77°C and 1.33 atm occupies a volume of 50.3 L. How many moles of the gas are present? (Hint: Since moles
Nina [5.8K]

The number of moles of the gas at 77°C and 1.33 atm occupies a volume of 50.3 L is 2.35 moles. It can found with the help of Ideal gas equation.

<h3>What is Ideal Gas equation ? </h3>

The ideal gas equation is formulated as : PV = nRT.

In this equation, P refers to the pressure of the ideal gas, V is the volume of the ideal gas, n is the total amount of ideal gas that is measured in terms of moles, R is the universal gas constant, and T is the temperature.

Given ;

  • Pressure = 1.33 atm
  • Volume = 50.3 ltr
  • Temperature = 77 (+273 k) = 350K

We know ;

  • Gas constant (R) = 0.081 L atm/mol K

Formula used ;

n = PV / RT

n = 1.33 x 50.3 / 0.081 x 350k

  =  2.35 moles.

Hence, The number of moles of the gas at 77°C and 1.33 atm occupies a volume of 50.3 L is 2.35 moles

Learn more about Ideal Gas here ;

brainly.com/question/27922399

#SPJ1

3 0
2 years ago
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