When thinking about breaking bonds and making bonds, you can use the abbreviation BENDOMEX, (Breaking bonds (B) is endothermic (ENDO) and (Making bonds (M) is exothermic (EX))
This channel has honestly saved my A levels….thank you so much. You make the explanations really easy and straightforward, which is something my teacher admittedly can’t do ❤
at 24:00 when finding the enthalpy change of neutralizing by dividing Q by the number of moles..do we usually calculate the number of moles of the acid or the base? why did we choose HCl in this case and not NaOH? I'm aware that in this example both have the same number of moles so it does not matter but what if they have different moles (ex: if we are using a strong acid with a weak base)? please reply, thanks!!!
@@leeshtan heyy after searching about it, the most sensible answer i got is that we will use the moles of the limiting reagent, so though id share what i found with u:)
At 43:23 while calculating mean bond enthalpy i don't get why we only write C(g) + 4H(g) + 4O(g) ,why are we not talking about CO2 and 2O2 ? Please clarify
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When thinking about breaking bonds and making bonds, you can use the abbreviation BENDOMEX, (Breaking bonds (B) is endothermic (ENDO) and (Making bonds (M) is exothermic (EX))
This channel has honestly saved my A levels….thank you so much. You make the explanations really easy and straightforward, which is something my teacher admittedly can’t do ❤
Thank you so much! This video has literally cleared all my confusion on the topic!! Especially Hess’s Cycle!
You're so welcome!
From south Africa here u honestly so Great at explaining and make sure not to leave anything out
I wish these videos were there for me when I gave my A-levels 😢
Bless you, Mr Harris. This has been beyond helpfull.
at 24:00 when finding the enthalpy change of neutralizing by dividing Q by the number of moles..do we usually calculate the number of moles of the acid or the base? why did we choose HCl in this case and not NaOH? I'm aware that in this example both have the same number of moles so it does not matter but what if they have different moles (ex: if we are using a strong acid with a weak base)? please reply, thanks!!!
I am not sure how accurate I am but I think you choose the more reactive compound
@@leeshtan heyy after searching about it, the most sensible answer i got is that we will use the moles of the limiting reagent, so though id share what i found with u:)
At 43:34 while calculating mean bond enthalpy why had u multiplied 4×413? For C-H bonds?
At 43:23 while calculating mean bond enthalpy i don't get why we only write C(g) + 4H(g) + 4O(g) ,why are we not talking about CO2 and 2O2 ?
Please clarify
Because we are writing the constituent elements of CO2 and 2O2 = C and 4O
35:52 i didnt really understand how they should add up to zero? Im quitw lost so could you explain more?
In the first route he got -732, and from the other route he got +732.. so when you add these together you get 0
39:01 why we do not need to take enthalpy of combustion of oxygen?
currently watching this 5 minutes before my mocks 🗣️🗣️🗣️
i wanted to confirm for bond enthalpy, do the arrows always face down or it depends.
For combustion arrows point down to combustion products and for formation they point upwards from elements in standard states. 👍
is this 9701 syllabus???
yup
yes
Yes