The given question is incomplete. The complete question is:
When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.
Answer: The vant hoff factor for sodium chloride in X is 1.9
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
= freezing point constant
i = vant hoff factor = 1 ( for non electrolyte)
m= molality =
Now Depression in freezing point for sodium chloride is given by:
= Depression in freezing point
= freezing point constant
m= molality = 
Thus vant hoff factor for sodium chloride in X is 1.9
Explanation:
1. Boyle's Law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
(At constant temperature and number of moles)
2. Charles' Law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.
(At constant pressure and number of moles
3. Gay Lussac's Law states that tempertaure is directly proportional to the pressure of the gas at constant volume and number of moles of gas
(At constant volume and number of moles)
Answer:
ΔG°rxn = -69.0 kJ
Explanation:
Let's consider the following thermochemical equation.
N₂O(g) + NO₂(g) → 3 NO(g) ΔG°rxn = -23.0 kJ
Since ΔG°rxn < 0, this reaction is exergonic, that is, 23.0 kJ of energy are released. The Gibbs free energy is an extensive property, meaning that it depends on the amount of matter. Then, if we multiply the amount of matter by 3 (by multiplying the stoichiometric coefficients by 3), the ΔG°rxn will also be tripled.
3 N₂O(g) + 3 NO₂(g) → 9 NO(g) ΔG°rxn = -69.0 kJ
Answer:
There is 2.52 kJ of energy released (option 4)
Explanation:
Step 1: Data given
The enthalpy of fusion of methanol (CH3OH) is 3.16 kJ/mol
Mass of methanol = 25.6 grams
Molar mass of methanol = 32.04 g/mol
Step 2: Calculate moles of methanol
Moles methanol = mass methanol / molar mass methanol
Moles methanol = 25.6 grams / 32.04 g/mol
Moles methanol = 0.799 moles
Step 3: Calculate energy transfer
Energy transfer = moles * enthalpy of fusion
Energy = 0.799 moles * 3.16 kJ/mol
Energy = 2.52 kJ released
There is 2.52 kJ of energy released
Answer: pH of resulting solution will be 13
Explanation:
pH is the measure of acidity or alkalinity of a solution.
Moles of 
ion = 
Moles of 
ion = 
For neutralization:
1 mole of ion will react with 1 mole of ion
0.01 mol of ion will react with =
of ion
Thus (0.012-0.01)= 0.002 moles of are left in 20 ml or 0.02 L of solution.
![[OH^-]=\frac{0.002}{0.02L}=0.1M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B0.002%7D%7B0.02L%7D%3D0.1M)
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![pOH=-log[0.1]=1](https://tex.z-dn.net/?f=pOH%3D-log%5B0.1%5D%3D1)
Thus the pH of resulting solution will be 13
