Also I have made more progress on the problem I think, if anyone is willing to take a look at my work:
In part (3) we are given 9.6 grams of Na(C2H5COO) which amounts to 0.0999349382 or roughly 0.1 moles. In one liter of solution we get 0.1M. Then using the Henderson-Hasselbalch equation...
Thank you, I'm always down to use easy solutions!
For point of reference though, what would the dissociation of the salt look like? How would that be written in this case?
Hi everyone. I'm doing the multi-step problem above, and I've found myself stuck at part 3.
For step 1, I determined that the equation that we're dealing with is:
$$CH_3CH_2CO_2H + H_2O ⇔ CH_3CH_2CO_2^- + H_3O^+$$
I've also determined that the molar mass of propanoic acid is about 74.08g/mol...
I think I'm starting to understand. My only other question is in regards to the transition from step 6 to step 7 in my picture. How does ##\frac x {d+x}## become ##1 - \frac d {d+x}##?
Okay so radius is a function of x. Q is a function of charge density and x. But charge density is a nested function within our charge function, and it varies differently than charge. Charge density varies in tandem with radius, as they are functions of the same input, which is the position along...
Okay tell me if I'm visualizing this correctly. Is it correct to say that since ## λ = \frac Q L ## and therefore ## λL = Q## then charge is going to vary with x in the same way as the radius (d + x) does? And that begets ##dq = λdx## where the small change in charge is accompanied by a similar...
So if our charge element was positioned at the leftmost end of our rod, it would be the distance (d + x) away from A. So we have ##V = \frac {KQ} {d + x}##. When we then think about dV, a small change in electric potential, it corresponds to a ratio of a small change in charge, dq, to a small...
Okay, so our small changes in length are occurring at the starting point d + x, and they are so negligible that we can essentially treat the whole thing as the static value d + x rather than as a changing value? That being said can't we just consider it a constant and factor it out in front of...
rude man, no worry about the re-edits! I agree, the use of d as a variable was a terrible choice, especially considering they already opted to use a capital L for the other distance variable. But they're getting paid to write physics books and I am not, so who I am to question their variable...
I'm still struggling with setting up this integral. So we have our original equation ## V = \frac {KQ} r ##. We can rewrite this as ## V = \frac {KλL} r## which becomes ## V = \frac {Kλx} r##. r for our point at A would be described by d + x so we write ## V = \frac {Kλx} {d + x}##. We know...
Homework Statement
"A rod of length L lies along the x-axis with its left end at the origin. It has a non-uniform charge density λ=αx where α is a positive constant. a) What are the units of α? b) Calculate the electric potential at A.
Homework Equations
Linear charge density: λ = Q/L where Q...