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Yes but the answer the exam board wants for our school is: neutralisation.

Great to hear it’s useful.

@@aylshamhighchemistry1353 thank so much

Good luck!!

Glad I could help

You should get a similar curve.

You're welcome

You're welcome!

For GCSE chemistry we simplify and only really look at the electronic configuration of the first 20 elements. None of these are in the transition metal block so we don’t really look at them. The videos on this channel only cover GCSE course content. As an aside some versions of the periodic table do have group numbers running for each column of the transition metals.

I’m sorry you’re struggling. It’s really difficult to answer this without seeing the question and what you’re doing with your calculator. The main mistake people make is not using the x10x button. If you’ve got a different question post it and I can help with the answer. If not then it’s worth getting your teacher to see what’s going on with your calculator.

Sorry about that. The aim is to try and work out the amount of oxygen in the air. In this experiment the oxygen gas reacts with the copper to form solid copper oxide. If we compare the gas volume before and after then we can make a conclusion about how much oxygen was in the air.

Take your time thank you 🙏

No problem. Pure substances melt and boil at set temperatures. Mixtures have a melting point range and a boiling point range. In the question you’ve written T and U have only one temperature for melting and boiling indicating that they are pure. Whereas substance V starts melting at 25 and finishes at 30oC. This is a range and it’s the same for boiling. Substance V is therefore a mixture. Hope that helps.

Wow! Thank you! But I thought if the numbers have a huge difference then it is impure and if it's close then it's pure? Does this mean that if it's a range it must be impure and if it's only one number then it's pure? (I am learning Secondary 3 Chemistry) 😅

Certainly for GCSE in England what you have stated is correct (I’m afraid I’m not familiar with Secondary 3 or what is required).

@@aylshamhighchemistry1353 I am in the 9th Grade so far and thank you so much for the help! Is it alright for me to ask you any other questions that I might have in the near future?

For GCSE you probably don’t need this level of detail. However, this might help. H2SO4 dissociates in water to form H+ and SO42- ions. A H+ ion will combine to a NH3 molecule to form NH4+. Finally NH4+ and SO42- join together by electrostatic forces to form ammonium sulfate. So,there is no chance to produce water The reaction is given below: NH3 + H2SO4 =( NH4)2SO4

@@aylshamhighchemistry1353 thanks but I was confused as NH3 is a gas when it reacts with acid it must be in its aqueous form right? in NH4OH as an alkali

No problem, if it helps I’ve just lifted the following extract from Wikipedia: ‘Dry, powdered ammonium sulfate may be formed by spraying sulfuric acid into a reaction chamber filled with ammonia gas.’

That’s very kind. Thank you.

The questions in the videos are not past paper questions. They are similar and there were for example questions asking candidates to calculate surface area:volume ratio in both 2023 and 2022 on the Edexcel papers.

I’m not sure you would write it like this. However, there have been a couple of GCSE exam questions where candidates are asked to do this. Hope that helps.

If I’m honest I haven’t taught A level for 20 years. Aylsham is 11-16. The basics at GCSE are true for A Level but they may well want a more detailed answer referencing van set Waals forces of attraction. Sorry I can’t be if more help.

@@aylshamhighchemistry1353 thank you so much

Glad it helped!

You deserve a medal for being subscribed to the most boring channel on RUclips for so long! I have fond memories of your year group, I hope uni is going well.

@aylshamhighchemistry1353 no its not boring! It helped me get this far didn't it?!

It’s great to hear that it’s useful. Thanks so much.

Great question! Mass number/Avogadro constant = mass (g) of 1 atom. This would be a very tiny number!

@@aylshamhighchemistry1353 can I ask why would it give me mass of one atom? This is where I got super confused! Thank you so much for the reply !

No problem. The Avogadro constant is the number of atoms needed to turn the mass number into the same number of grams. For example the mass number of carbon is 12 so if you have 6.02x10^23 atoms of carbon this will have a mass of 12g. Therefore if you divide the mass number by the Avogadro constant it will give you the mass of 1 atom. For example, the mass number of carbon is 12. Divide this by 6.02x10^23. It will give 1.99 x 10^-23 g. Hope that helps - it is a bit complex.

@@aylshamhighchemistry1353 thank you so much! I now see your point! It makes so much more sense now!

That’s great. Glad it helped.

Glad it was helpful!

You're welcome!

@@aylshamhighchemistry1353 Do you have any more in-depth videos on how to "create" these reactions?

Sorry for delay in reply. I’ve only just seen this. All the questions with video links are in the workbook 4 on this page. aylshamhighchemistry.com/workbooks/ Hope that helps.

It’s in the year 11 workbook. Download here: aylshamhighchemistry.com/workbooks/

@@aylshamhighchemistry1353 Thank you!

Very pleased it was helpful.

Glad it helped

Glad it helped!

Glad it helped!

Glad it was helpful!

S2O32-(aq) + 2H+(aq) ---> S(s) + SO2(g) + H2O(l) (Can’t do subscripts for formula numbers or superscripts for ion charges on my phone)

Aylsham High Chemistry

Yes, you can write the unit that way.

Richard Lock Thanks for the ‘before present’ definition, I wasn’t aware of that one. Regards Phil Brockington.