Excellent lesson. Studying chemistry for one year on an Access course, and once again Mr Rintoul's channel has been outstanding. I hope I can find use in this channel once I'm at Uni! Thanks again!
I'm studying biomedical science, really similar to biochemistry. I know the modules can be different, so I'll break down what I covered on Access: - Physics (pass/fail) and Advanced Physics (graded) - Environmental Science (graded, I think?) and Ecology (graded) - Foundation chemistry (graded) and redox, equilibria and energetics (graded) - Loads of graded biology content, but cannot for the life of me remember the unit names haha. - Maths (ungraded) and study skills (ungraded) If you've covered anything around these, I'm sure they'll be of ample information. It also depends on what you're issue is concerning; the likelihood of being accepted or how you'll find the content; - De Montfort accepted me with just passes; though I got 45/45 distinctions (I had low expectations :P) - they even offered me a £1000 scholarship for enrolling with an Access qualification. So universities are keen on giving you a chance :) - In regards to what foundational knowledge you will need is anyones guess, but you will often find the universities will go right back to basics to ensure you're ready to grasp the better stuff. (We were literally taught that the mitochondria was the power house of the cell). Don't fret too hard, and remember that there will be handfuls of other people with Btecs, Access and other non-A-level qualifications in the same boat as you. If you're worried about whether you are capable, please take heed - I left A-levels with E's in everything (that I didn't get U's in) and have managed to attain Firsts in EVERYTHING thus far at Uni. I hope you do really well, and please ask any more questions. :)
Joshua Macleod thanks man, motivating words. I'm heading towards my last exams and ontrack for 45/45. I'll have to look into the scholarship at Nottingham. How did you go about getting that? was it the same route as scholarships for AAA students. My plan is to read over A2 topics and bring up my maths over summer.
+E Rintoul We have done learning all of CHEM 4 and start learning CHEM 5 after the half term in January. We started CHEM 4 straight after are AS exams were over last year so that's why we've finished now.
+E Rintoul No, we haven't done any CHEM5 at this point, so no doubt your videos will really help a lot. Are teachers described CHEM5 as "it either makes you or breaks you" so this is going to be interesting....
hppsgm Haha! Yeah, CHEM5 can be a bit of a killer BUT if you can do well on it, it's a really great chance to get loads of UMS as the grade boundaries are normally quite low!
hi, if anyone can reply..why did you get methanol as 0.07mol when it states that 0.17mol was found in equilibrium mixture.. i want to know why he had to do 0.24-0.17 process... and also i want to know what is 0.24mol, 0.17mol, 0.07 mol separately to clarify my doubt... :(
Funny how the practise question that we learnt on the video came in one of my class tests and now I’m watching the video after the test, as Gordon Ramsay would say “FUCK ME!!!!!”
Don't suppose there is a pH of weak acids calculations/ pH calculations involving neutralising video(s) coming soon? Happy New Year! Ps. Your videos are saving my grades right now!! Thank you! :)
Its 0.07 and 0.24bc: The mole ratio is given in the equation: 1 mole of CO reacts with 2 moles of hydrogen t make 1 mole of methanol If you start off with 0.24 mol of CO and there are 0.17 mol at equilibrium then 0.24 - 0.17 mol of CO have reacted. From the stoichiometry of the equation this makes the same number of moles of methanol. From the stoichiometry of the equation twice of much hydrogen must have reacted. Subtract moles of hydrogen reacted from original moles present (given in the question) to get moles of hydrogen remaining. You now have the number of moles of each component.
@@sherrypahilanga2172 a year late but screw it. I’m no expert (which is why I watched this video) but I’d assume 0.240 was the initial mol number and 0.170 is the mol after equilibrium was reached. Therefore the mol difference is just before and after so 0.240 - 0.170
I really appreciate your videos as they are really helpful. I've had to learn most of the unit 4 content on my own. The teacher that got appointed in September didn't have a proper grasp of the A-Level content and didn't 'teach'. The management have only recently just got rid of her. I'm a bit worried as I feel as though we're behind with things as we haven't even touched Unit 5.Thankfully, having looked at your videos and doing past papers questions has reassured me. Anyways, I was wondering if you'd do any EMPA practice papers?
+Ahmodul Kabir Having looked at a few comments recently, there are a lot of schools that work through CHEM4 first and then move onto CHEM5. We do it differently where I'm at and as such, I have almost finished CHEM 5 with one class and with the other, we are a good way through CHEM4. I would say that by this point you probably want to be about done with CHEM4 so that you have plenty of time for CHEM5. However, you can go quite quickly through CHEM5 if you work hard at it!
Its great that you're still covering the old spec Sir. Also, i have uni offers from york and manchester to study chemisty, which should i make my first choice?
+SorleyBoy98 Excellent, well done! That all depends on the offers! The highest offer should go above the other otherwise you stand a chance of not getting a place if you miss you top choice. But, if they are both the same offer, it depends on which uni you like best! I love Manchester as a place and York is lovely too, if a little quieter.
i am doing something to do with quadratic formulas in our chemical equilibria topic, could you explain what its used for please? Love the videos by the way, recommended them to all my friends
was getting confused at 13:21 but then noticed i did it right. (You said 0.07) but in reality its 0.17 therefore it would be a 1:2 ratio x2 = 0.34 0.38 - 0.34 = 0.04 for H and 0.17 for Methanol? Correct?
At 14:24 Q3a where has the hydrogen concentration gone from the methanol concentration (0.14)? Because it's says 0.07. Do you not add the change in concentration for both reactants to get the concentration of the product? (0.21) . I'm confused.
Sir, at minute 10:00 you say their is 1.5 moles of B. My question is why didn't you put 1.5 as your coefficient in regards to the B. Because then you wouldn't have 0.4 moles of B (like you did with A) but rather you would have 0.8 as it is now a 1:2 ratio rather than a 1:3 ratio. Can you please clarify this, thank you.
+"So where are you going?" (chapter 81 verse 26) I'm not sure that you're really understanding what is going on here... The numbers that I wrote down are nothing to do with the ratio, they area the actual values of those reactants and products before and at equilibrium. The ratio can NEVER be 1:2 as the numbers in the equation do not allow this. The ratio of A:C and of B:C is 1:3. ALWAYS. You mention having 0.4 moles of B. That isn't the case, rather, from the amount of C at equilibrium and from the ratio of B to C, we know that 0.4 moles of B must have REACTED, leaving 1.10 moles (1.50 - 0.400). Does that make any more sense now? If not, let me know how I can help further.
Thanks for the vids, when you say temperature is the only condition effecting kc why doesn't pressure effect it? as surely the equilibrium will shift depending on number of moles therefore changeingthe amounts of reactants and products
Sebastian Owen i think it's because the equilibrium constant is only a measure of concentration so although there are moles the constant won't change by it.dunno why temperature changes it
Because pressure has the same effect as concentration, its the change of the amount of molecules in a given volume therefore would not have an effect on Kc because the pressure of the molecules on the opposite reaction would change too
Sir, Ples may you me reply me? @14:04 i don'nt understand why you subtracted 0.380 and 0.14. Isn't the question asking about how much hydrogen is used to come to an equibrium?
+MaleVip We know that 0.07 moles of carbon monoxide reacted in reaching equilibrium. Because the ratio of carbon monoxide to hydrogen is 1:2, that means that 0.14 moles of hydrogen must have reacted. Because there were initially 0.38 moles of hydrogen, the amount at equilibrium would be 0.38-0.14. Get it?
i will try and explain this: so, an increase in the concentration of products (or a decrease in the reactants) will cause Kc to increase. but a decrease in the concentration of products (or a increase in the reactants) will cause Kc to decrease. you should also note that you can find out if the products/reactants are being increased/decreased using Le Chateliers. hope this helped xx
@@areebaa1595 conc doesn't have an effect on Kc value... What you're saying is correct form equilibrium, howevever, temp does have an effect the Kc value but only temp
Excuse me sir, but the question at 12:00 is confusing me. If the hydrogen lost 0.14mols then surely the methanol will have gained those moles aswel as the moles lost from the carbon monoxide? totalling at 0.21 moles of ethanol?
@@joana359 do you know why before equilibrium, there are 2.5 moles in total of reactants and products but then at equilibrium (11:31) there are 2.9 moles in total? How does that make sense?
+Nía Feisal As that is for the H2 and there is 2 moles of H2 in the equation so the concentration of H2 (in the equation for Kc) is squared. But the one next to it (which is the Concentration of CO) is not squared as there is only one mole of CO in the equation so the concentration of CO is to the power of 1 (so not squared). Hope this helps!!
