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xeze [42]
2 years ago
9

A fuel was burned for 5 min, increasing the temperature of 10.0 g of water with a density of 1.00 g/ml by 9.0 oC. The fuel relea

sed _____ ca
Chemistry
1 answer:
jekas [21]2 years ago
7 0

The fuel released 90 calories of heat.

Let suppose that water experiments an entirely <em>sensible</em> heating. Hence, the heat released by the fuel is equal to the heat <em>absorbed</em> by the water because of principle of energy conservation. The heat <em>released</em> by the fuel is expressed by the following formula:

Q = m\cdot c \cdot \Delta T (1)

Where:

  • m - Mass of the sample, in grams.
  • c - Specific heat of water, in calories per gram-degree Celsius.
  • \Delta T - Temperature change, in degrees Celsius.

If we know that m = 10\,g, c = 1\,\frac{cal}{g\cdot ^{\circ}C} and \Delta T = 9\,^{\circ}C, then the heat released by the fuel is:

Q = (10\,g)\cdot \left(1\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (9\,^{\circ}C)

The fuel released 90 calories of heat.

We kindly invite to check this question on sensible heat: brainly.com/question/11325154

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2 years ago
What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?
jasenka [17]

Answer:

71.372 g or 0.7 moles

Explanation:

We are given;

  • Moles of Aluminium is 1.40 mol
  • Moles of Oxygen 1.35 mol

We are required to determine the theoretical yield of Aluminium oxide

The equation for the reaction between Aluminium and Oxygen is given by;

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.

Therefore;

1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen

1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium

Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.

4 moles of aluminium reacts to generate 2 moles aluminium oxide.

Therefore;

Mole ratio Al : Al₂O₃ is 4 : 2

Thus;

Moles of Al₂O₃ = Moles of Al × 0.5

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                         = 0.7 moles

But; 1 mole of Al₂O₃ = 101.96 g/mol

Thus;

Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol

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C

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