How many grams of chlorine gas can be produced from the decomposition of 73.4

1 answer:

6 0

1) first, we have to convert the grams to moles of AuCl3 using the molar mass of the molecule.

molar mass of AuCl3= 197 + (35.5 x 3)= 303.5 g/mol

73.4 g (1 mol AuCl3/ 303.5 g)= 0.242 moles

2) now. let's convert moles of AuCl3 to moles of chlorine gas (Cl2) using the mole-mole ratio

0.242 mol AuCl3 (3 mol Cl2/ 2 mol AuCl3)= 0.363 mol Cl2

3) finally, we convert moles to grams using the molar mass of Cl2.

molar mass of Cl2 = 35.5 x 2= 71.0 g/mol

0.363 mol Cl2 ( 71.0 g/ 1 mol)= 25.8 grams

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Arrange the bromine-containing compounds CaBr2, Br2, NBr3, CBr4 in order of increasing predicted boiling point.

Fudgin [204]

Explanation:

In an ionic compound, there will be strong force of attraction between the combining atoms due to the opposite charges present on them.

For example, $CaBr_{2}$ is an ionic compound where calcium has a +2 charge and each bromine atom has a -1 charge.

Therefore, in order to break the bond we need to provide more heat. Hence, boiling point of calcium bromide will be the highest.

$NBr_{3}$ is a covalent compound and as nitrogen is more electronegative in nature and also has a lone pair of electron hence, there will be a net dipole moment.

$CBr_{4}$ is also a covalent compound. And, as bromine is more electronegative than carbon atom so, dipole moment is in the outwards direction. Hence, in $CBr_{4}$ there will be zero dipole moment.