Explanation:
As the given reaction is as follows.
So, according to the balanced equation, it can be seen that rate of formation of 
will be twice the rate of disappearance of 
.
And, it is known that rate of disappearance of reactant will be negative and rate of formation of products will be positive value.
This means that,
Rate of the reaction = -Rate of disappearance of
= ![k[H_{2}S][Cl_{2}]](https://tex.z-dn.net/?f=k%5BH_%7B2%7DS%5D%5BCl_%7B2%7D%5D)
= 
= 
M/s
Therefore, calculate the rate of formation of as follows.
Rate of formation of = 
= 
M/s
Thus, we can conclude that the rate of formation of is M/s.
At STP, also known as standard temperature and pressure, 1 mole of a gas occupies 22.4 L. Since we are given with the volume of 6.3L, we calculate the amount of gas in mol.
n = (6.3L)/ (22.4L/mol) = 0.28125 mol
We are given with the mass of 6.7 g. Therefore, the molar mass or molecular weight of the gas is equal to,
6.7g/0.28125 mol = 23.82 g/mol
Answer:
O FX will be greater than FY
Explanation:
<em>Surface tension</em> can be defined as the force required to stretch one film of a given fase (usually with liquids).
This required force is proportional to the liquid's surface tension. This means that the higher the surface tension, the higher the required force to stretch it is.
The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
