Question

on heating 0.56g of copper [ll] Bromide decomposed to give 0.36g of copper [l] bromide and some bromine.How many moles of bromine molecules were given off.

Answer 1

-0.0009 moles of bromine molecules were given off.

To determine the number of moles of bromine molecules that were given off during the decomposition of copper [II] bromide, you need to know the molar masses of copper [II] bromide and copper [I] bromide, as well as the mass of copper [I] bromide produced during the decomposition.

The molar mass of copper [II] bromide is 223.66 g/mol, and the molar mass of copper [I] bromide is 200.07 g/mol. The mass of copper [I] bromide produced during the decomposition is 0.36 g.

To determine the number of moles of bromine molecules that were given off, you can use the following steps:

Calculate the mass of copper [II] bromide that was decomposed by subtracting the mass of copper [I] bromide produced from the original mass of copper [II] bromide: 0.56 g - 0.36 g = 0.20 g

Calculate the number of moles of copper [II] bromide that was decomposed by dividing the mass by the molar mass: 0.20 g / 223.66 g/mol = 0.00089 moles

Calculate the number of moles of copper [I] bromide produced by dividing the mass by the molar mass: 0.36 g / 200.07 g/mol = 0.00179 moles

Calculate the number of moles of bromine molecules that were given off by subtracting the number of moles of copper [I] bromide from the number of moles of copper [II] bromide: 0.00089 moles - 0.00179 moles = -0.0009 moles

Note that the number of moles of bromine molecules that were given off is negative because the mass of copper [I] bromide produced during the decomposition is greater than the mass of copper [II] bromide that was decomposed. This means that some of the bromine molecules that were given off during the decomposition of copper [II] bromide were used to form the copper [I] bromide that was produced.

The final answer is -0.0009 moles.

Answer 2

To find the number of moles of bromine molecules that were given off, you need to know the atomic masses of copper and bromine. The atomic mass of copper is 63.55 g/mol and the atomic mass of bromine is 79.904 g/mol.

First, you need to calculate the number of moles of copper bromide that was decomposed. You can do this by dividing the mass of copper bromide by its atomic mass: 0.56 g / (63.55 g/mol + 79.904 g/mol) = 0.01 mol.

Next, you can use the balanced chemical equation for the decomposition of copper bromide to determine the ratio of bromine molecules to copper bromide molecules. The balanced chemical equation is:

CuBr2 -> Cu + Br2

This equation shows that for every one molecule of copper bromide that decomposes, two molecules of bromine are produced.

Finally, you can use the number of moles of copper bromide and the ratio of bromine molecules to copper bromide molecules to calculate the number of moles of bromine. In this case, there are 0.01 mol of copper bromide and 2 molecules of bromine are produced for every 1 molecule of copper bromide, so the number of moles of bromine is 0.01 mol * 2 molecules / 1 molecule = 0.02 mol.

Therefore, 0.02 moles of bromine molecules were given off when 0.56 g of copper bromide decomposed.

Step-by-step explanation: