What is the common uses of bases

Will NaBr, CO2, Ca3(PO4), or KZS raise a liquid's boiling point the most when

Vladimir79 [104]

Answer:

$K_2S$

Explanation:

According to the boiling point elevation law described by the equation $\Delta T_b = iK_bb$ , the increase in boiling point is directly proportional to the van 't Hoff factor.

The van 't Hoff factor for nonelectrolytes is 1, while for ionic substances, it is equal to the number of moles of ions produced when 1 mole of salt dissolves.

$NaBr$ would produce 2 moles of ions per 1 mole of dissolved substance, sodium and bromide ions.

$Ca_3(PO_4)_2$ is insoluble in water, so it would barely dissociate and wouldn't practically change the boiling point.

$K_2S$ would dissociate into 3 moles of ions per 1 mole of substance, two potassium cations and one sulfide anion.

$CO_2$ is a gas, it would form some amount of carbonic acid when dissolved, however, carbonic acid is molecular and would yield i value of i = 1.

Therefore, potassium sulfide would raise a liquid's boiling point the most if all concentrations are equal.

6 0

3 years ago

If the solubility of a gas is 10.5 g/L at 525 kPa pressure, what is the solubility of the gas when the pressure is 225 kPa? Show

Talja [164]

Answer:

4.5 g/L.

Explanation:

To solve this problem, we must mention Henry's law.

Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

It can be expressed as: P = KS,

P is the partial pressure of the gas above the solution.

K is the Henry's law constant,

S is the solubility of the gas.

At two different pressures, we have two different solubilities of the gas.

<em>∴ P₁S₂ = P₂S₁.</em>

P₁ = 525.0 kPa & S₁ = 10.5 g/L.

P₂ = 225.0 kPa & S₂ = ??? g/L.

∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.

8 0

3 years ago

Molarity to percent by mass. Convert 1.672 mol/L MgCl2(aq) solution to percent by mass of MgCl2 in the solution. The solution de

nignag [31]

Answer:

$\%m/m=14\%$

Explanation:

Hello!

In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:

$[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}$