The freezing point of water is 0.00°C at 1 atmosphere.
If 13.40 grams of ammonium sulfide, (68.20 g/mol), are dissolved in 153.2 grams of water. The molality of the solution is m.
The freezing point of the solution is °C.

Molality is: 1,283m

Freezing point of the solution is: -7,159°C.

Explanation:

Molality is defined as moles of solute per kg of solvent

Moles of ammonium sulfide (solute):

13,40g × (1mol / 68,20g) = 0,1965moles

Molality is:

0,1965moles / 0,1532kg Water = 1,283m

The freezing point depression formula is:

ΔT = Kf × m × i

Where ΔT is change in temperature, k is solvent constant (1.86 °C/m for water), m is molality (1,283m) and i is Van't Hoff factor (3 for ammonium sulfide -van't hoff factor is the number of ions which is dissociates in the solution, for (NH₄)₂SO₄ ions are 3-. Replacing:

ΔT = 1,86°C/m × 1,283m × 3

ΔT = 7,159°C

As freezing point of water is 0°C, freezing point of the solution is 0°C - 7,159°C = -7,159°C

I hope it helps!

i'm assuming that the correct answer is indeed the first one/ "a"..

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-