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When one pound of propane is combusted, 22.8 MJ of energy is released. Determine the mass of propane (in grams) required to produce 547 kJ of energy.

Respuesta :

sebassandin

Answer:

[tex]m=10.9g[/tex]

Explanation:

Hello!

In this case, since the heat released per pound of propane is:

[tex]\frac{22.8MJ}{1lb}[/tex]

If 547 kJ of energy is produced, the following mathematical set up provide the required mass of propane (in grams) to do so:

[tex]m=547kJ*\frac{1MJ}{1000kJ} *\frac{1lb}{22.8MJ} *\frac{453.6g}{1lb}\\\\ m=10.9g[/tex]

Best regards!

Eduard22sly

The mass of propane (in grams) required to produce 547 KJ of energy is 10.88 g

Conversion scale

1 pound = 453.592 g

1 MJ = 10⁶ J

1 KJ = 10³ J

Data obtained from the question

  • 22.8 MJ of energy = 1 pound of propane
  • 547 kJ of energy =?

22.8×10⁶ J of energy = 453.592 g of propane

Therefore,

547×10³ J = (547×10³ × 453.592) / 22.8×10⁶

547×10³ J = 10.88 g of propane

Thus, 10.88 g of propane is required to produce 547 KJ of energy

Learn more about energy released in a reaction:

https://brainly.com/question/25141046

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