OK, so using the value of x to be 1.53
H2 = F2= 0.47
2HF= 5.06
if X= 2.69
H2=F2= -0.69
2HF = 7.38
Oh so THATS what you meant. Sorry my bad :P yeah one of the values get -ve so that will be discarded, and the other will be the correct value. Thanks.
And what do you think about Q5?
Doesn't the question only asks for the final concentration of HF only? Plus that's what I tried doing, but got confused by the value of 'x'.
PS: This is not my homework, but I am preparing for the finals, so this is the past paper with only the answers, so showing me the correct way would only...
yeah i know, it would be (2 + 2x), but like I said which value of x should I use, seeing that both are possible. Anyways I don't get the right answer by using either value :/
i dont, i just followed the standard procedure and got 2 positive values (ususally in this type of cases i get 1 -ve and +ve so i can discard the -ve value), which can't be right, but cannot figure out my mistake :/
Oh right right you mean
Ksp= (1.2x10^-3)(2.4x10^-3)^2
and then equate this to the final solubilities. Thanks.
And what is my mistake with questions 4 and 5?
Hi I have attached the method I tried to solve these questions, along with additional question. I am not sure what I did wrong, but the quadratic equations i get give out two completely possible values of the variable. The question and the my tries are labelled respectively. Thanks
Here is the original question. As for the ICE table, I am not too good with it, can only perform the basic generic operations. Would be helpful if you could show me and I will learn from it.
Homework Statement
H2(g) + S(s) = H2S(g) Kc= 6.8x10^-2 If 0.2 moles of H2 and 1.0 mole of S are heated in a 1L vessel upto 90C, what will be the partial pressure of H2S at equilibrium? Can someone help me with this step by step?Homework Equations
The Attempt at a Solution
Kc = 6.8X10^-2 = x /...