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DENIUS [597]
3 years ago
13

Three hundred grams of aluminum are heated from −50 ◦c to 300 ◦c. if the specific heat of aluminum is 0.88 j/(g ◦c), what is the

change in entropy of the aluminum?
Chemistry
1 answer:
enyata [817]3 years ago
4 0
Q = mCΔT
Q is heat in Joules, m is mass, C is specific heat, and ΔT is change in temp

Q = (300g)(.88)(350 degrees) = 92400 J or 92.4 kJ
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Answer:

The answer is A

Explanation:

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A gas occupies a volume at 34.2 mL at a temperature of 15.0 C and a pressure of 800.0 torr. What will be the volume of this gas
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The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
     P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
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3 0
3 years ago
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What’s the mass of 2.80 mol of ca
Ludmilka [50]

Answer:

112.22 g

Explanation:

mass = no. of moles x atomic mass

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If the initial temperature of a movable cylinder was 50 degrees Celsius
slega [8]

Answer:

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Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

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323 × V₂ = 10 × 273

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Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

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