0.528344 gallons in a 2 liter
Answer:
Question 1 is 21.0g
Explanation:
50 x 4.18 x 3.1 / 0.444 / 69.6= 21.0
Answer:
See below
Explanation:
propane mole weight = 44 gm / mole
100 gm / 44 gm / mole = 2.27 moles
From the equation, 5 times as many moles of OXYGEN (O2)are required
= 11.36 moles of oxygen
at <u>STP</u> this is 254.55 liters of O2 (because 22.4 L = one mole) and
Using oxygen as 21 percent of air means that
.21 x = 254.55 = x = <u>1212.12 liters of air required </u>
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
Explanation:
The two ways that energy can be transferred are by doing work and by heat transfer.
