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
Answer:
A group of cells ready to form roots a stem and the first leaves
Correct Question: what is the oxidizing agent in the reaction.
2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)
Answer: MnO4-is the oxidizing agent
Explanation:
In the reaction 2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)
Oxidizing agent oxidizes other molecules while the themselves get reduced.
oxidizing agents give away Oxygen to other compounds.
MnO4-is the oxidizing agent because
On the reactants side
Oxidation number of Mn in 2MnO4- is +7
Oxidation number of Cl- is -1
On the products side
Oxidation number of Mn is +2
While oxidation number of Cl is zero
Therefore the oxidizing agent is 2MnO4 because is oxidizes Chlorine from -1 to 0 while itself got reduced from oxidation state of +7 to +2
Answer: 0.002 M
Explanation:
The balanced chemical equation for ionization of 
in water is:
According to stoichiometry :
1 mole of ionizes to give =2 mole of 
ions
0.001 mole of ionizes to give = 
mole of ions
Thus of a solution of 0.001 M aqueous sulfuric acid is 0.002 M
The answer is 27.92g . working is shown in the picture above.
