Answer:
Vapour pressure of benzene over the solution is 253 torr
Explanation:
According to Raoult's law for a mixture of two liquid component A and B-
vapour pressure of a component (A) in solution = 
vapour pressure of a component (B) in solution = 
Where
are mole fraction of component A and B in solution respectively
are vapour pressure of pure A and pure B respectively
Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr
So, vapour pressure of benzene in solution = 
= 253 torr
Answer:
The correct statements are:
The rate of disappearance of B is twice the rate of appearance of C.
Explanation:
Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.
3A + 2B → C + 2D
Rate of the reaction:
![R=-\frac{1}{3}\times \frac{d[A]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=R%3D-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{1}\times \frac{d[C]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D)
![-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{2}\times \frac{d[D]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D)
The rate of disappearance of B is twice the rate of appearance of C.
![\frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![2\times \frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{1}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=2%5Ctimes%20%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
Answer:o It is important to realise that mixing will be small unless there are electrons in the 4a1 LUMO, this is why NH3 is pyramidal while BH3 is planar! ... This mixing is very strong and stabilises the 3a1 MO substantially and hence NH3 is trigonal pyramidal and not planar.
Explanation: