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Naily [24]
3 years ago
8

Pentane is a small liquid hydrocarbon, between propane and gasoline in size at C5H12. The density of pentane is 0.626 g/mL and i

ts enthalpy of combustion is ΔH°comb = –3535 kJ/mol. Estimate the fuel value and the fuel density of pentane.
Chemistry
1 answer:
Goshia [24]3 years ago
4 0

Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and 3.073\times 10^4kJ/L respectively.

Explanation :

Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).

As we are given that:

\Delta H^o_{comb}=-3535kJ/mol

Fuel value = \frac{\Delta H^o_{comb}}{\text{Molar mass of pentane}}

Molar mass of pentane = 72 g/mol

Fuel value = \frac{3535kJ/mol}{72g/mol}

Fuel value = 49.09 kJ/g

Now we have to calculate the fuel density of pentane.

Fuel density = Fuel value × Density

Fuel density = (49.09 kJ/g) × (0.626g/mL)

Fuel density = 30.73 kJ/mL = 3.073\times 10^4kJ/L

Thus, the fuel density of pentane is 3.073\times 10^4kJ/L

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Answer is: A) The solution turns blue litmus to red.

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3 years ago
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The specific heat of copper is 0.40 joules/ g °c. How much heat is needed in joules to change the temperature of a 55 gram subst
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Answer : The amount of heat needed is, 1188 J

Explanation :

Formula used :

q=m\times c\times (T_2-T_1)

where,

q = heat needed = ?

m = mass of copper = 55 g

c = specific heat capacity of copper = 0.40J/g^oC

T_1 = initial temperature = 20.0^oC

T_2 = final temperature = 74.0^oC

Now put all the given values in the above formula, we get:

q=55g\times 0.40J/g^oC\times (74.0-20.0)^oC

q=1188J

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Answer:

Q = 4.056 J

Explanation:

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∴ m = 406.0 mg = 0.406 g

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∴ T1 = 33.5°C ≅ 306.5 K

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⇒ Q = (0.406 g)(1.85 J/gK)(5.4 K)

⇒ Q = 4.056 J

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