Answer: The temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg is 920 K
Explanation:
According to ideal gas equation:
P = pressure of gas = 2300 mm Hg = 3.02 atm (760mmHg=1atm)
V = Volume of gas = 15 L
n = number of moles = 0.6
R = gas constant =
T =temperature = ?
Thus the temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg is 920 K
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
Since it’s asking for chemical to thermal, B.
Answer:
2
Explanation:
The coefficient for O is 2 and this is an example of a combustion reaction. With the help of the coefficient 2 infront of oxygen, this equation now demonstrates law of conservation of mass.
