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
increases my factor of 10
Answer:
d
. Sc2O5
Explanation:
Hello,
In this case, when forming oxides from a metal and oxygen, for us to find out each element's subscript, we must exchange them as shown below, considering +5 for scandium:
For that reason, the answer is d
. Sc2O5
Best regards.
Noble gases react very unwillingly, because the outermost shell of electrons orbiting the nucleus is full, giving these gases no incentive to swap electrons with other elements. As a result, there are very few compounds made with noble gases. Like its noble gas comrades, neon is odorless and colorless.
Answer:- 13.6 L
Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.
Standard pressure is 1 atm that is 101.325 Kpa.
Boyle's law equation is:
From given information:-
= 58.7 Kpa
= 23.5 L
= 101.325 Kpa
= ?
Let's plug in the values and solve it for final volume.
On rearranging the equation for
= 13.6 L
So, the volume of hydrogen gas at STP for the given information is 13.6 L.
