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Concentrated aquous sulphuric acid is 98% H2SO4 (w/v) and has a density of 1.8gmL^-1. Molarity of the solution .
(1)1M
(2)1.8M
(3)10M
(4)1.5M
Mole fraction of solute in aqueous solution of 30% NaOH.
(1).16
(2).05
(3).25
(4).95
Deduce the SI units for the gas constant, R.
how to calculate the exact mass of dueterium.the exact mass is 2.0141017779.how to calcute the atomic mass of the isotopes of other elements?
EQUILIBRIUM
1. At What extent (1-alpha) can be considered negligible?

BONDING
1.Does hybridised orbital mostly form sigma-bond?
2.Please explain me inorganic RESONANCE,as I am finding it hard to choose the correct option in the questions of STABILITY ?
3. How to do the MOT diagram for polyatomic molecules?
4. What is Synergic bond?
LogKp/Kc+log RT=0 is a relationship for the reaction
(1)PCl5=PCl3+Cl2
(2)2SO2+O2=2SO3
(3)H2+I2=2HI
(4)N2+3H2=2NH3
For the ques :- PH of 10 * 10^(-8) of HCL.. the solution given is


[H+] total = [H+] acid + [H+] water
Since HCl is a strong acid and is completely ionized
[H+] HCl = 1.0 x 10-8
The concentration of H+ from ionization is equal to the [OH-] from water,
[H+] H2O = [OH-] H2O
= x (say)
[H+] total = 1.0 x 10-8 + x
But
[H+] [OH-] = 1.0 x 10-14
(1.0 x 10-8 + x) (x) = 1.0 x 10-14
X2 + 10-8 x – 10-14 = 0
Solving for x, we get x = 9.5 x 10-8
Therefore,
[H+] = 1.0 x 10-8 + 9.5 x 10-8
= 10.5 x 10-8
= 1.05 x 10-7
pH = – log [H+] = – log (1.05 x 10-7) = 6.98

so, x= [OH-] =9.5 X 10-8, so, POH is 7.02

But If it is so what would be the case for 10*-7 M of HCL, x = [OH-] would be 1.6 X 10-7
and POH <7, BUT HOW IT IS POSSIBLE as it cant be less than 7 and PH-POH=PH would be more than 7 . HOW it is possible?
Two vanderwaal gases A and B are at corresponding state.the critical temperature and pressure of gases are
Pc/atm. Tc/K
A. 48. 150
B. 33. 125
Find the volume of B at this corresponding state if the volume of A is 1.5L.
find out the ph of pure water
what is the maximum limit of addition of acid and base to prevent the change of pH
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