Honestly the best chemistry youtuber ever, no one even comes close. Never stop making these videos please!!!

@chemistrytutor

4 ай бұрын

That's really lovely to hear! Thank you for your kind words 😊

Omg I was waiting for this! I couldn’t find a detailed enough video on this topic and was worried but then just the next day, I see that you’ve uploaded a video on this exact topic. Thank you so much.

@chemistrytutor

Жыл бұрын

You're very welcome! It's lovely to know it's useful for you 😊

I can't thank you enough for all your videos! You make everything complicated demystified!

@chemistrytutor

4 ай бұрын

I'm really pleased you think so 😊

This is so great! I had to pause the video to say thank you, I really appreciate your help!!

@chemistrytutor

Жыл бұрын

That's really lovely to hear 😊 ... don't forget to finish the video 😁

the red- ox thing is revolutionary, thanks!

@chemistrytutor

Жыл бұрын

Such a small detail, but so useful 👌

I am really grateful for all of your videos! My finals are knocking at the door but I still have lots of confusions in Chemistry. Your videos give me hope that I can do well, regardless of the short period of time remaining.

@chemistrytutor

Жыл бұрын

Thank you! That's really lovely to know they're useful. Good luck for the exams! 👍

The best explanation of redox out there. Thanks!!!!

@chemistrytutor

2 ай бұрын

That's really nice feedback, thank you 😊

best chem teacher out there. ngl. Thanks A LOT

@chemistrytutor

6 ай бұрын

Thanks for your lovely feedback 😀

I remember using you to pass my high school exams a few years ago, I'm back to refresh my knowledge.

@chemistrytutor

Ай бұрын

Welcome back 🙏 Hope things are going well 😃

the explanation in this video was really good! thank u so much sir !!! 😊

@chemistrytutor

2 ай бұрын

I really appreciate the feedback! Very nice to know it's useful 👍

Thank you so much for this video!!!

@chemistrytutor

8 ай бұрын

You're very welcome! Glad it's useful 😊

You're such a king #lifesaver

@chemistrytutor

6 ай бұрын

Thanks 😀

Hi sir very helpful video, could you make a video explaining redox and electrode potentials please I dont understand what we need need to know and apply

@chemistrytutor

10 ай бұрын

Thanks for the feedback 😀 That topic is on my to do list for this term. In the meantime I have made a few questions walkthroughs so you can see the application of the topic. Here are the links: kzread.info/dash/bejne/iJ2bktWrm5WYY8Y.htmlfeature=shared kzread.info/dash/bejne/poatl9KpeJzgg7Q.htmlfeature=shared kzread.info/dash/bejne/maKExKqdZ5uznaw.htmlfeature=shared

19:05 is solved as elimination mthd for quadratic eqtn in maths, for those that don't understand

@chemistrytutor

Ай бұрын

😀

Your videos are a great help! Do you consider doing ligand substitution?

@chemistrytutor

Жыл бұрын

Thanks for the feedback! I'm moving on to question walkthroughs for a bit after the next video. I'll make a ligand substitution one for that 😀

@RingsideReels

Жыл бұрын

@@chemistrytutor Thanks, Sir

Bro I was just searching up redox reactions 😂

@chemistrytutor

Жыл бұрын

This is good timing!! 😀

wow, thank you so much.

@chemistrytutor

Жыл бұрын

Very welcome 😀

This guy is the goat

@chemistrytutor

9 ай бұрын

😀👍

Brilliant!!!!

@chemistrytutor

5 ай бұрын

Thank you 😊

THANK YOU SO MUCH!

@chemistrytutor

4 ай бұрын

You're very welcome 😀 👍

hi, when we're working out O.S do we ignore the coefficient at the front e.g. in 2MnO4^- the O.S of Mn is +7 so we dont include the 2 at the front right ? ty!

@chemistrytutor

4 ай бұрын

That's correct 👌 The coefficient doubles everything, so you'd just be working it all out twice. And the O.S is for a single element, so 2Cl^- has an OS of -1 (there are just two of them)

When the reaction takes place in alkali conditions I understand that you are supposed to use OH- ions but I am unsure how you are supposed to balance the equation and all the practice questions I have done I have been incorrect. Could you possibly make a short video to cover this? Also this video was insanely helpful so thanks sooo much

@chemistrytutor

2 ай бұрын

Thanks for the feedback. I'm really pleased it's useful! I'll add that video to my list 😃

Hi, I don't fully understand how the charge of the LHS at 17:23 is +7? Could you please explain?

@chemistrytutor

Жыл бұрын

Each of the H+ ions has a 1+ charge. The MnO4^1- has a 1- charge. The sum of those numbers is 7+ or +7 😀

Thank you so much ❤

@chemistrytutor

3 ай бұрын

Very welcome 🙏

hi, i don't understand how the charge at 22:34 is +7 for 2MnO4^-1 and not 7.5? this was my working: (2Mn) + (8x-2) = -1 the 8 is for the 4oxygens x big 2 (2Mn) + (-16) = -1 2Mn = 15 Mn = 15/2 Mn = 7.5

@chemistrytutor

Жыл бұрын

You just work it out for one single MnO4^1- you don't need to do it for two. Although, if you do, you get the same answer. You just went astray because you doubled all of the atoms but not the charge for the ion

can u plz make a video on carbonyl compounds?

@chemistrytutor

Жыл бұрын

Good idea 💡 I'm going to make a video about the halogens next. But I'll do at least one carbonyl chemistry video, and other organic 2nd year videos after that 😀

22:33 Sir I’m so confused on how Mn had a charge of +7 Do we work that out based on the fact that after writing the equation Mn had a charge of 2+ and since it’s after it gained 5 electrons then it must’ve originally had a charge of +7?

@chemistrytutor

Ай бұрын

It's a tough one. But I think it's easier to know that oxygen is -2 and work it from there rather than thinking about the full equation. I go through it earlier as a worked example at 10:04 in the video 😃

Sorry this may sound really stupid but at 15:00 how would you know what side of the reaction to add the electrons to as for iodine i would have have put the two electrons on the right hand side was zero however that wouldnt work as it wouldnt make 0 so is there a way i can know what side to put the electrons pelase?

@chemistrytutor

Жыл бұрын

Good question, there are four ways: 1) you are adding something negative to one side in order to make both sides have the same charge... so add them to the side that is currently more positive (or less negative) 2) if you know the oxidation State change e.g. +2 --> 0 then you can work out where electrons should go. So in this example 2e- must be added on the left as reduction is occurring. 3. If you know the type of process reduction or oxidation that tells you the side to add them red-ox ... if its red the e go on the left, oxidation they're on the right. 4. If you put them on one side for one half equation, the second one you're combining with it must have e on the other side!

@MariamNuhu-fc9vz

Жыл бұрын

Words can't describe how much you have helped me so far i will surely be using your videos a lot this summer your amazing thank you sir!

@chemistrytutor

Жыл бұрын

@@MariamNuhu-fc9vz thanks for your kind words. Best of luck with your studying!

Do we need to know about disproportionation reactions for AQA?

@chemistrytutor

7 ай бұрын

Yes definitely. I've covered it in a halogen video as well kzread.info/dash/bejne/ZJmCvNWEncvaptI.html

16:39 Sir at d LHS we had (1-) and (+1) which is supposed to b zero(0) so why adding the e-

@chemistrytutor

Ай бұрын

We needed 2 H+ to balance the H from the H2O, meaning the LHS was 1+ overall before adding the e-

@IyunolaSasi

Ай бұрын

@@chemistrytutor sir I already understand now. In fact I did d last example my self and it was accurate 😊😁. I'll continue watching now 🔰. Thanks for ur attention sir

Thanks!!!!!

@chemistrytutor

4 ай бұрын

😀👍

Sir, do u have KZread videos on physics topics???

@chemistrytutor

Ай бұрын

I'm afraid I'm currently only doing chemistry videos

@IyunolaSasi

Ай бұрын

@@chemistrytutor OK Sir

15:31 sorry if this is stupid but why not just multiply No3 by 2 and NO2 by 3?

@chemistrytutor

6 ай бұрын

If you did it like that the nitrogen wouldn't balance. Your equation would be suggesting that you had created a nitrogen from somewhere. Additionally charge wouldn't be conserved as NO3 has a 1- charge and NO2 is neutral

@bigg.grizzlybear2670

5 ай бұрын

@@chemistrytutor Hello i have a question. In the half eqn of H2O that turns it into H2 and OH(-), I added an H+ ion and 2 electrons to the left to form H2O + h(+) + 2e ----> H2 + OH(-) but its wrong. Where did i go wrong?

@chemistrytutor

5 ай бұрын

@bigg.grizzlybear2670 Tough question! This is a bit of a weird scenario that doesn't follow the standard rules. A lot of redox takes place in acidic solution, hence adding H+ ions to one side. On this occasion we make OH- ions (base/alkali), so the alkali won't be being made from the acid. We need to balance this by using 2H2O and therefore we need to make 2OH- ions. Then we add the electrons to balance the charge or to address the decreasing oxidation States (whichever way you prefer to think about it)

@swodeshsingh5442

5 ай бұрын

​@@chemistrytutor can you elaborate on why it gets formed 0:03

@chemistrytutor

5 ай бұрын

@swodeshsingh5442 hi. I'm not sure exactly what you mean... has some autocorrect happened?

@ 8:11, Why can't the Hydrogen change to +2 each instead of oxygen changing? so 4 + - 4 = 0

@chemistrytutor

4 ай бұрын

Hydrogen only has 1 electron and 1 proton. So its highest oxidation State is only +1

@dwcLDN

4 ай бұрын

@@chemistrytutor Thanks for the explanation 👍

@chemistrytutor

4 ай бұрын

@@dwcLDN 👌

Hello Sir, when can we expect a video on halogens?

@chemistrytutor

Жыл бұрын

Realistically, a week or so. Not before the end of next weekend 😀

@toxins5803

Жыл бұрын

@@chemistrytutor Amazing to hear Sir, will you be doing A2 videos next?

@chemistrytutor

Жыл бұрын

@@toxins5803 exactly 😀 Halogens is the last AS video to make

@chemistrytutor

Жыл бұрын

Here you go... kzread.info/dash/bejne/jGtmqLh9qbeuoaw.html 😀

I don't understand the tricky part😥😥

@chemistrytutor

Ай бұрын

Which bit do you mean?

@IyunolaSasi

Ай бұрын

@@chemistrytutor 24:06 everything about the tricky REDOX

@chemistrytutor

Ай бұрын

@IyunolaSasi this is showing that sometimes we need to simplify more than the electrons. We find the common multiple of electrons and remove them. Then we need to check for H+ and H2O (and anything really but these are the common ones) and see what repeats on both sides. A chemical equation should only show what changes so if we start with 10H+ and finish with 6H+ for instance, only 4 of those H+ have actually changed. To fix this we remove 6H+ from both sides