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3 years ago

how many kilojoules of heat are absorbed when 0.38 grams of chloroethane vaporizes at its normal boiling point?

Chemistry

2 answers:

frozen [14]3 years ago

8 0

Answer:

1.6 ×10⁻¹ kj

Explanation:

AnnZ [28]3 years ago

4 0

Answer:

1.6 ×10⁻¹ kj

Explanation:

Given data:

Mass of chloroethane = 0.38 g

Heat absorbed = ?

Solution:

Molar mass of chloroethane = 64.51 g/mol

Molar heat of vaporization = 26.4 kj/mol

Number of moles:

Number of moles = mass/ molar mass

Number of moles = 0.38 g/ 64.51 g/mol

Number of moles = 0.006 mol

Heat absorbed:

0.006 mol × 26.4 kj/mol

0.16 kj

or 1.6 ×10⁻¹ kj

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The burning of magnesium is a highly exothermic reaction. How many kilojoules of heat are released when 0. 75 mol of Mg burn in

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Taking into account the definition of enthalpy of a chemical reaction, the quantity of heat released when 0.75 moles of Mg are burned is 451.5 kJ.

<h3>Enthalpy of a chemical reaction</h3>

The enthalpy of a chemical reaction is known as the heat absorbed or released in a chemical reaction when it occurs at constant pressure. That is, the heat of reaction is the energy that is released or absorbed when chemicals are transformed into a chemical reaction.

The enthalpy is an extensive property, that is, it depends on the amount of matter present.

<h3>Heat released in this case</h3>

In this case, the balanced reaction is:

2 Mg(s) + O₂ (g) → 2 MgO(s) + 1204 kJ

This equation indicates that when 2 moles of Mg reacts with 1 mole of O₂, 1204 kJ of heat is released.

When 0.75 moles of Mg are burned, then you can apply the following rule of three: if 2 moles of Mg releases 1204 kJ of heat, 0.75 moles of Mg releases how much heat?

$heat=\frac{0.75 moles of Mgx1204 kJ}{2 moles of Mg}$