That salt hydrolysis formula right? Yeah you can.
For question 8 use van der Waals equation bro. Just put b=0 as they said that volume occupied by CO2 is negligible.
So bro( P + an^2 \over V^2) (V-n b ) = n R T
Now putting n =1 and b=0 i get
(P + a \over V^2 ) V = R T
Now in the given question i have to find P but so a ,V, T should be known but only a and T are given please help @Rudra_d
Hm didn’t notice that. I have an exam today. Will try after that.
@Achyut_Singh dekho jo tum bol rahe woh sahi hai
Uske aage mai batata hu
Dekho v me quadratic banao kuki 1 particular value of t and p be 1 hi value aayega v ka to jo tum quadratic banaoge uska discriminant 0 kar dena
.ans dikh jayega
Apka din magalmay ho🙃
But why D = 0 . What is the reason behind it @Vibhu_Kumar please explain
Equal roots me kya lagate ho
Toh tumhe kaise pata hai ki yaha par equal roots ki condition lagegi ? Mujhe kuch samjh nhi aa rha h ek bar samjhado please
Koi 1 temp aur pressure ki condition par 1hi value na aaygi volume ki aur quadratic hamlog 2 value de dega
Par dono value 1 hi honi chahiye na isliye equal roots jaisa condition hoga to hamlog D=0 Istamal kardenge
@Vibhu_Kumar bhai ye 4,6 bhi batado bhagwan tumhara bhala karega
Baad me dekhenge abhi kaam hai thora
is the answer to que no.4 is four ?
At first equivalence point, NaHCO3 will be present which is an amphiprotic species.
Use pH= (pka1 + pka2)/2 to get pH1 which will be 6.3
At II equivalence point which is the end point of titration there will only be Na2CO3 which will cause hydrolysis in 2 steps but 2nd step hydolysis will be negligible.
[CO3]2- will come as 0.04 M or 1/25 M by balancing the equivalents. then calculate first hydrolysis constant. use Kh1=Kw/Ka2 it will come 10^-6.
Find [OH-] and pOH. So pH2 comes out to be 10.3
this will give (pH2-pH1) = 4
Yes it is the correct answer, I checked from the key just now
bro write as
C_6H_6 + 3H_2 -> C_6 H_12
x y –
- y -3x x
earlier x + y = 60
then y - 2x = 30
Bhai yeh line samjh nhi aayi