Recent content by devon

You've definitely done more than enough, no need to do any more reading! I'll just have a look, I'm pretty sure the K out matches the Na in but I'll have another read to be sure. You're a life saver, cheers :)

Thanks so much for all of the help! Yeah I was just being lazy with the rounding because typing everything out on here takes ages. It's crazy how a 1000 fold smaller change in concentration can cause such a large shift in potential difference. Just to make sure, I didn't have to do anything with...

So is it: 1.875x10^11 Na+ ions per m/6.02x10^23 divided by avagadros number to go back to moles = 3.12x10^-13 moles per m. As 1 meter = volume we figured out earlier divide this number by the axon volume (divide it by 1000l first as we want to compare it to the 15milimole/litre). So 3.12x10^-13...

Homework Statement Calculate the number of sodium ions entering a non-myelinated axon during the action potential, per metre of axon length. The change in potential is 100 mV, the axon membrane capacitance per unit length is 3 × 10−7 F/m, and the charge on an ion is 1.6 × 10−19 C. What...

Thank you so much for your help, I'll definitely make sure to look at the actual chemistry behind these reactions! So I assume because you didn't say anything that my answer for 2 1-phenylcyclohexanol (b) is correct?

Ok so I did a bit of study and got a better understanding of the topic. Bromide addition reactions that aren't aqueous (the tetrachloride isn't used, just to identify that it's not in water) "break" the double bond and replace either side of it with Bromide ions. So for 1. I would go with C as 1...

I get either 856kPa or 8.4 atm as well so I guess I'm done as long as ~8.7atm = 8.4atm! Thanks so much for all of the help!

I figured out the correct n down the bottom of this thread, I just divided the ion concentration by avagadro's number to get 0.3321155762mol/L. So do I have to divide my final answer with the volume of 1 blood cell? I'm getting a bit confused.

But doesn't the volume being divided on the bottom give me a vastly different answer? This is how the equation is given to me in my textbook: Do I need to divide my final answer with the volume of the blood cell I found in my first attempt?

Thank you so much for all of your help and putting up with my 1 million questions!

Ah its LkPa/kmol. So that would make my final answer 856.0262371 kPa?

Ah that makes complete sense! So I just need to divide that number by avogadro's number and then do all the conversion? (2x10^23ion/L)/6.022x10^23 = 0.3321155762mol/L? Osmotic Pressure = 0.3321155762mol/L x (8.3145LPa/Kmol) x 310K = 856.0262371 Pa?

Oh so am I assuming that the ion concentration given is actually mol/m^3? That makes way more sense. So I convert the m^3 to L to get mol/L (2x10^26/1000L) = 2x10^23mol/L. So the answer would be: Osmotic Pressure = (2x10^23mol/L) x (8.3145LPa/Kmol) x 310K = 5.15499x10^26 Pa? I assume that the...

Is this the equation I'm after? The units for the impermeat ion's concentration are throwing me of (m^3). M is the molar concentration of dissolved species (units of mol/L). R is the ideal gas constant (0.08206 L atm mol-1 K-1, or other values depending on the pressure units). T is the...

Homework Statement i) A human erythrocyte (red blood cell) may be approximated as a disc around 2 microns in height and with a diameter of 10 microns. If the pressure inside were to become large enough for the cell membrane to rupture, where would you expect it to fail, assuming the membrane...