If the reaction is represented by:
PCl₃ + Cl₂ <-> PCl₅ (exothermic)
the mole fraction of chlorine in the equilibrium mixture will change according to the following:
Decrease the volume: decrease
Increase the temperature: increase
Increase the volume: increase
Decrease the temperature: decrease
It’s the BOA not the dog or kangaroo
Answer:
The empirical formula of compound is C₂H₆O.
Explanation:
Given data:
Mass of carbon = 12 g
Mass of hydrogen = 3 g
Mass of oxygen = 8 g
Empirical formula of compound = ?
Solution:
First of all we will calculate the gram atom of each elements.
no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms
no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms
no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms
Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.
C:H:O = 1/0.5 : 3/0.5 : 0.5/0.5
C:H:O = 2 : 6 : 1
The empirical formula of compound will be C₂H₆O
solution:
Weight of caffeine is W = 0.170 gm.
Volume of water is V= 10 ml
Volume of methylene chloride which extracted caffeine is v= 5ml
No of portions n=3
Distribution co-efficient= 4.6
Total amount of caffeine that can be unextracted is given by
![w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\=0.170[\frac{46}{46+5}]^3\\=0.170[\frac{46}{51}]^3\\=0.170[\frac{97336}{132651}]\\=0.170\times0.734=0.125gms](https://tex.z-dn.net/?f=w_%7Bn%7D%3Dw%5Ctimes%5B%5Cfrac%7Bk_%7BDx%7Dv%7D%7Bk_%7BDx%7Dv%2Bv%7D%5D%5En%5C%5C%3C%2Fp%3E%3Cp%3Ew_%7B3%7D%3D0.170%5B%5Cfrac%7B4.6%5Ctimes10%7D%7B%284.6%5Ctimes10%2B5%29%7D%5D%5E3%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5B%5Cfrac%7B46%7D%7B46%2B5%7D%5D%5E3%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5B%5Cfrac%7B46%7D%7B51%7D%5D%5E3%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5B%5Cfrac%7B97336%7D%7B132651%7D%5D%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5Ctimes0.734%3D0.125gms)
amount of caffeine un extracted is 0.125gms
amount of caffeine extracted=0.170-0.125
=0.045 gms
Here we will use the general formula of Nernst equation:
Ecell = E°Cell - [(RT/nF)] *㏑Q
when E cell is cell potential at non - standard state conditions
E°Cell is standard state cell potential = - 0.87 V
and R is a constant = 8.314 J/mol K
and T is the temperature in Kelvin = 73 + 273 = 346 K
and F is Faraday's constant = 96485 C/mole
and n is the number of moles of electron transferred in the reaction=2
and Q is the reaction quotient for the reaction
SO42-2(aq) + 4H+(aq) +2Br-(aq) ↔ Br2(aq) + SO2(g) +2H2O(l)
so by substitution :
0 = -0.87 - [(8.314*346K)/(2* 96485)*㏑Q → solve for Q
∴ Q = 4.5 x 10^-26
