The chemical bonds in CH4 are all single bonds. C only can bond 4 times because it needs 8 electrons in it's outer shell and only has four right now. The bonds represented are all single bonds because there are two electrons present on each side of the carbon. Two electrons, in this case, equals one bond.
PH = -log [H3O+]
4.15 = -log [H3O+]
[H3O+] = 10^(-4.15)
[H3O+]= 7.08 × 10^-5
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: -
When bismuth-212 undergoes alpha decay, it becomes ²⁰⁸Tl
Explanation: -
Mass number of ²¹²Bi = 212
Atomic number of ²¹²Bi = 83
When alpha decay occurs the mass number decreases by 4 and the atomic number decreases by 2.
Mass number of daughter = 212 - 4 = 208
Atomic number of daughter = 83 - 2 = 81
It is the atomic number of Thallium Tl.
Thus the daughter nucleide is ²⁰⁸Tl.
Answer:
When sodium chloride dissolves in water to make a saturated solution there is a 2.5 per cent reduction in volume. ... The solubility of salt does not change much with temperature, so there is little profit in using hot water.