What is the formula of hexaaquamanganese(ii) sulfate? express your answer as a chemical formula?

Chemistry

1 answer:

Digiron [165]3 years ago

3 0

Answer : The formula of hexaaquamanganese(II)sulfate is $[Mn(H_2O)_6]SO_4$ .

Explanation :

Rules for writing formulas of coordination complexes :

Central metal is written first and then the ligands are written along with their prefixes.

The anion ligands are written before the neutral ligands.

If there are more than one anion ligands or neutral ligands then they are written in alphabetical order.

The given name of complex is, hexaaquamanganese(ii) sulfate.

In this, the central metal is manganese which is written as Mn and the ligand is water (which is a neutral ligand) is written as $H_2O$ and an anion is sulfate which is written as $SO_4$ .

Water is a neutral ligand and sulphate has a charge of -2. As the overall complex is neutral the metal bears a +2 charge which is represented in Roman numeral after the metal.

Therefore, the formula of hexaaquamanganese(II)sulfate is written as $[Mn(H_2O)_6]SO_4$ .

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2 years ago

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$