The mass of a substance per unit volume is the substance's density.
D = m/v.
Answer:
[H_3 O^+] = 1.0 ×10^-13
Explanation:
If we multiply the left side we get, 1e - 13. We add 13 to the right side while subtracting the remaining value from the left side (H3O+) than combine like terms. As you will get pH = 13.00
Answer:
Sample A is a mixture
Sample B is a mixture
Explanation:
For sample A, we are told that the originally yellow solid was dissolved and we obtained an orange powder at the bottom of the beaker. Subsequently, only about 30.0 g of solid was recovered out of the 50.0g of solid dissolved. This implies that the solid is not pure and must be a mixture. The other components of the mixture must have remained in solution accounting for the loss in mass of solid obtained.
For sample B, we are told that boiling started at 66.2°C and continued until 76.0°C. The implication of this is that B must be a mixture since it boils over a range of temperatures. Pure substances have a sharp boiling point.
Answer:
Boiling point of the solution is 100.78°C
Explanation:
This is about colligative properties.
First of all, we need to calculate molality from the freezing point depression.
ΔT = Kf . m . i
As the solute is nonelectrolyte, i = 1
0°C - (-2.79°C) = 1.86 °C/m . m . 1
2.79°C / 1.86 m/°C = 1.5 m
Now, we go to the boiling point elevation
ΔT = Kb . m . i
Final T° - 100°C = 0.52 °C/m . 1.5m . 1
Final T° = 0.52 °C/m . 1.5m . 1 + 100°C → 100.78°C
