P=0.0902 g/l
v=22.4 l/mol (stp)
M=vp
M=22.4 l/mol * 0.0902 g/l=2.020 g/mol
M=2.020 g/mol
382.85 Celsius is the temperature does 0.750 moles of an ideal gas occupy a volume of 35.9 L at 114 kPa.
Explanation:
Given data:
number of moles of the gas = 0.75 moles
volume of the gas = 35.9 liters
pressure of the gas = 114 KPa or 1.125 atm
R = 0.0821 latm/moleK
temperature of the gas T = ?
The equation used to calculate temperature from above data is ideal gas law equation.
the equation is :
PV = nRT
T = 
Putting the values in the above rewritten equation:
T = 
T = 655.9 K
To convert kelvin into celsius, formula used is
K = 273.15+ C
putting the values in the equation
C = 656 - 273.15
= 382.85 Celsius
Answer:
The enthalpy of the solution is -35.9 kJ/mol
Explanation:
<u>Step 1:</u> Data given
Mass of lithiumchloride = 3.00 grams
Volume of water = 100 mL
Change in temperature = 6.09 °C
<u>Step 2:</u> Calculate mass of water
Mass of water = 1g/mL * 100 mL = 100 grams
<u>Step 3:</u> Calculate heat
q = m*c*ΔT
with m = the mass of water = 100 grams
with c = the heat capacity = 4.184 J/g°C
with ΔT = the chgange in temperature = 6.09 °C
q = 100 grams * 4.184 J/g°C * 6.09 °C
q =2548.1 J
<u>Step 4:</u> Calculate moles lithiumchloride
Moles LiCl = mass LiCl / Molar mass LiCl
Moles LiCl = 3 grams / 42.394 g/mol
Moles LiCl = 0.071 moles
<u>Step 5:</u> Calculate enthalpy of solution
ΔH = 2548.1 J /0.071 moles
ΔH = 35888.7 J/mol = 35.9 kJ/mol (negative because it's exothermic)
The enthalpy of the solution is -35.9 kJ/mol
Separate Salt and Water Using Distillation
If you want to collect the water, you can use distillation. This works because salt has a much higher boiling point than water. One way to separate salt and water at home is to boil the salt water in a pot with a lid.
hope this helps!!
