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Triss [41]
2 years ago
6

A truck uses 6 kg of fuel which undergoes complete combustion. The heat produced during the complete combustion was measured to

be 220,000 kJ. Calculate the calorific value of the fuel.
Chemistry
1 answer:
Ludmilka [50]2 years ago
8 0

Answer:  Total mass of fuel = 4.5 kg Total heat produced = 180,000 kJ Heat produced by burning 1 kg of fuel = 180,000 kJ/4.5 kg = 40,000 kJ/kg. So, calorific value of fuel = 40,000 kJ/kg.

Explanation:

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D. With the same number of protons and different number of neutrons.
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What are some factors that might cause our percent yield to be greater than 100%? What are some factors that might cause it to b
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What are families 3-12 on the periodic table called
fredd [130]

Answer:

transition elements

Explanation:

6 0
3 years ago
Read 2 more answers
Acetylene, c2h2, has a standard enthalpy of formation, δh° = 226.7 kj/mol, and a standard entropy change for its formation from
VLD [36.1K]
ΔG⁰ = ΔH⁰ - T ΔS⁰

ΔG⁰ : Standard free energy of formation of acetylene

ΔH⁰ : Standard enthalpy of formation (226.7 kJ/mol)

ΔS⁰ : Standard entropy change (58.8 J / K. mol)

T : Temperature 25°C = 298 K (room temperature)

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5 0
3 years ago
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A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )
Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g)   (1)

The reaction rate of equation (1) is given by:

rate = k*[O_{3}][NO]     (2)

<u>We have:</u>

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s  

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

\frac{\Delta[NO_{2}]}{\Delta t} = rate

\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!                                

5 0
3 years ago
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