The total mass of the products is 10.76 g + 204.44 g = 215.20 g.
The masses of all the reactants but one are known so,
215.20 g - 120.00 g - 8.15 g - 75.00 g = 12.05 g
12.05 g is the mass of the unweighed barium nitrate.
Answer:
Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of 
to 1 kg of water.
Explanation:
1) Moles of NaCl ,
Mass of water = m= 1 kg = 1000 g
Moles of water = 
Vapor pressure of the solution = 
Vapor pressure of the pure solvent that is water = 
Mole fraction of solute(NaCl)= 
The vapor pressure for the NaCl solution at 17.19 Torr.
2) Moles of sucrose ,
Mass of water = m = 1 kg = 1000 g
Moles of water =
Vapor pressure of the solution = 
Vapor pressure of the pure solvent that is water =
Mole fraction of solute ( glucose)=
The vapor pressure for the glucose solution at 17.19 Torr.
p = p' = 17.19 Torr
Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of to 1 kg of water.
I think convergent but could be wrong
Answer:
2. 181.25 K.
3. 0.04 atm.
Explanation:
2. Determination of the temperature.
Number of mole (n) = 2.1 moles
Pressure (P) = 1.25 atm
Volume (V) = 25 L
Gas constant (R) = 0.0821 atm.L/Kmol
Temperature (T) =?
The temperature can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
1.25 × 25 = 2.1 × 0.0821 × T
31.25 = 0.17241 × T
Divide both side by 0.17241
T = 31.25 / 0.17241
T = 181.25 K
Thus, the temperature is 181.25 K.
3. Determination of the pressure.
Number of mole (n) = 10 moles
Volume (V) = 5000 L
Temperature (T) = –10 °C = –10 °C + 273 = 263 K
Gas constant (R) = 0.0821 atm.L/Kmol
Pressure (P) =?
The pressure can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
P × 5000 = 10 × 0.0821 × 263
P × 5000 = 215.923
Divide both side by 5000
P = 215.923 / 5000
P = 0.04 atm
Thus, the pressure is 0.04 atm
I really don’t know. Sorry
