If the heat of combustion for a specific compound is −1160.0 kJ/mol and its molar mass is 86.47 g/mol, how many grams of this compound must you burn to release 541.20 kJ of heat?

Answers

Answer 1

Answer:

40.34 g

Explanation:

First, we divide the heat to release by the heat of combustion to obtain the required moles of compound:

541.20 kJ/(1160.00 kJ/mol) = 0.4665 mol

So, we have to burn approximately 0.47 mol of the compound. We convert the moles to mass in grams by using the molar mass:

mass = molar mass x moles = 86.47 g/mol x 0.4665 mol = 40.34 g

Therefore, you must burn 40.34 grams of the compound to release 541.20 kJ of heat.


Related Questions

Base your answer on the information below. The hydrocarbon 2-methylpropane reacts with iodine as represented by the balanced equation below. At standard pressure, the boiling point of 2-methylpropane is lower than the boiling point of 2-iodo-2-methylpropane. Explain the difference in the boiling points of 2-methylpropane and 2-iodo-2-methylpropane in terms of both molecular polarity and intermolecular forces.

Answers

Answer:

See explanation

Explanation:

The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.

2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.

As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than  2-methylpropane.

what type of reaction is Au2S+H2---> 2Au+H2S

Answers

Answer:

single replacement reaction is type of reaction is Au2S+H2---> 2Au+H2S.

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