Recent content by Hunt_

Yes you plot P versus V and then at the point of inflection you set the 1st and 2nd derivatives to zero. This is where the critical pt exists. When you only have an equation and no graph whatsoever, you work out the derivatives and see if the critical pt exists. If you're not sure about your...

You are right. You can prove that H3PO4 third dissocitation is negligible as long as H3PO4 is the only source of hydronium ions in solution. ( either generally or u can take a numerical example and try it out yourself . ) But notice that the reactions u posted : H3PO4 + NaOH ---> NaH2PO4...

Thanks , I think it makes much more sense now.

When you write dW=Fdx you already assume the force is constant . Otherwise , why would F be outside the differential ? We should write dW = d(Fx) keeping F inside the differential as it is not a constant. dW = Fdx is correct only when F = constant. True ? Notice , when calculating the work...

Thanks for the notice, chemisttree. Wont happen again

In the case of an undamped oscillator, the work done by the system is written as ( assume initial position is 0 ) : W = - \Delta U = - K \frac{x^2}{2} But to verify this , we must assume that the force acting on the oscillator is constant , which is not true as F = f(x) according to...

I don't think there's any simple yet mathematically rigorous way of finding which geometry yields the more stable molecule , at least not that I know of.

p is a subshell that has 3 different orbitals. They are designated p_x , p_y , and p_z because each one is aligned along a specific axis in space. Each orbital can hold up to two electrons. While 5p subshell can hold up to 6 electrons , 5px / 5py / 5pz orbital can contain 2 electrons at most...

γ is the activity coefficient.

Your method is correct but you have few calculation errors. correct 4 - 4.76 = - 0.76 NOT 0.76 Where did you get 5.75 from ? You take the anti-log of -0.76 : 10^(-0.76} = [0.2 V NaOH/(0.5 V acetic acid - 0.2 V NaOH)] the correct expression should be : V (NaOH) = 0.37 V...

I will look later and see where you made your mistake . Sorry but there's No time now...

The reaction occurring is : NaOH + HOAC \rightarrow H_2 O + NaOAC For part 1 : If n_o is the initial amount of the acid , and x is the amount of NaOH. Then after the rxn is complete, n_o - x is the remaining amount of the acid and x is the amoun formed of its conjugate base. Notice...

correct Ofcourse it is wrong. The concentration of B^{4-} at equilibrium is [4x] and not x

You have an initial conc of HOAC and NaOAC ( both 0.1 M ) but both volumes are unknown. The HH equation says that pH = pK_a + log ( [base] / [acid] ) where [base] and [acid] are the concentrations are equlibrium. So what did you do ? You plugged in 0.1 for each , which is wrong. That is...

Sodium acetate is nothing but the conjugate base of acetic acid. According to HH equation, \frac{[NaOAC]}{[HOAC]} = 10^{pH - pK_a} At equilibrium , [NaOAC] = \frac{[NaOAC]_o V_{NaOAC}}{V_T} = \frac{(0.1)V_{NaOAC}}{V_T} [HOAC] = \frac{[HOAC]_o V_{HOAC}}{V_T} =...