Answer:
(4) 230 kPa
Explanation:
The temperature is constant, so the only variables are pressure and volume.
We can use Boyle’s Law.
Divide both sides of the equation by 
.
Answer:
They are clumped together and fluffy looking
Explanation:
I’ve seen what they looked like in a project during highschool we swabbed some books and door handles and saw which had the most bacteria. It was surprisingly the books
Answer:
T°fussion of solution is -18°C
Explanation:
We have to involve two colligative properties to solve this. Let's imagine that the solute is non electrolytic, so i = 1
First of all, we apply boiling point elevation
ΔT = Kb . m . i
ΔT = Boiling T° of solution - Boiling T° of pure solvent
Kb = ebuliloscopic constant
105°C - 100° = 0.512 °C kg/mol . m . 1
5°C / 0.512 °C mol/kg = m
9.7 mol/kg = m
Now that we have the molality we can apply, the Freezing point depression.
ΔT = Kf . m . i
Kf = cryoscopic constant
0° - (T°fussion of solution) = 1.86 °C/m . 9.76 m . 1
- (1.86°C /m . 9.7 m) = T°fussion of solution
- 18°C = T°fussion of solution
Solubility of barium chloride at 30 degree Celsius is 38.2g /100 g water and solubility of barium chloride at 60 degree Celsius is 46.6 g / 100 g water.
The quantity of barium chloride that is dissolved in water at 30 degree Celsius = 38.2 * [150/100] = 57.30 g.
The quantity of barium chloride that will be dissolved in water at 60 degree Celsius = 46.6 * [150/100] = 69.90 g
The difference between these quantities is the amount of barium chloride that can be dissolved by heating the barium chloride to 60 degree Celsius.
69.90 - 57.30 = 12.60 g. Therefore, 12.60 g of barium chloride can still be dissolved in the water by heating the water to 60 degree Celsius.
