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This is actually a question about Markownikofs rule (there is a video explaining it on my channel). The bromine adds to the terminal carbon because the carbocation intermediate is more stable when the positive charge is situated on the middle carbon. Both alkyl groups bonded to this C can donate electron density to stabilise. In the other scenario the bromine would add to the middle carbon and the positive charge would be sat on the terminal carbon which is less stable.

@@CrunchChemistrythank u so much ❤

in part (v)? one of the reactants in this redox system is iodide ions from the saturated solution the student was thinking of using to soak the strip oof filter paper in when making the salt bridge - be careful that we are talking about iodIDE ions as the reactant as iodINE is the product of the oxidation. In using NaI for the salt bridge we would have three redox reactions going on - oxidation of V2+, oxidation of I- and reduction of VO2+!

Absolutely, which topics specifically? AS or A level?

@@CrunchChemistry AS Chemistry, Thank you!

That was certainly the case a few years ago but interestingly I'm seeing exam questions on the roll-on effect of human error on experimental results much more often now.

Which part of the video are you referring too?

@@CrunchChemistry polymer and monomer.the qtn of combining monomers forming polymer

While you were forming polymer,those carbons

@@PalsFilms can you give me the time in the video? With more complicated monomers it is a case of making sure we know the difference between the carbons either side of the C=C double bond which are the ones that will form the polymer chain backbone, and any other carbons in the monomer which will become side groups.

Thanks for replying i solved it i wasn't paying attention to the given properly

Because the ratio of iodine to aluminium is 3:2 according to the balanced equation. The number of mol of iodine is 3/2 greater than that of aluminium so this is easiest way to scale up.

Place the 0 on your ruler on the starting pencil line and if your spot is more of a splodge measure to the half way point - either way as long as you use the same method for each measurement it doesn’t matter as you are calculating a ratio

The Nernst equation is describing the relationship between E, the non-standard electrode potential, and the concentration of ions (the oxidised species) in a half cell. In a half cell (e.g. Cu^2+ + 2e- -> Cu) the reduced species is always the metal electrode which is a solid, and since the concentration a solid is constant it is taken to be simply 1. You can revise this here ruclips.net/video/WXcI-j6xjhA/видео.html 😊

Thank you 😊! Indeed that would complete the picture but at A level Gibbs energy is treated very superficially, if at all, which is frustrating but hence the focus on the individual variables.

This model of ionic compounds dissolving assumes that once the ions are hydrated then they are electrically neutral and there is no connection between them, whether they are sodium or magnesium or chloride ions. There is a limit to how much of any solute we can dissolve in a solvent, including water, and it is dependent on a complex interplay of enthalpy and entropy factors ... ruclips.net/video/hMQguvQNN3U/видео.html We also should be aware that introducing common ions into a solution has an effect on the solubility of a solute (common ion effect) ...ruclips.net/video/CaMRiPhw6Q0/видео.html Adding other ions may also cause existing ions that are in solution to precipitate out ... ruclips.net/video/HpqfivWf22w/видео.html Overall, it is really messy and complicated picture - the best we can do is to appreciate all the factors involved and then consider each mixture individually. Great question! Hope this helps!

Thanks for the feedback!

Sorry, I've only just seen this! I'd need to see the whole question but you've probably figured it out now.

You can use either assuming you have the concentration and volume to calculate the number of moles (work through the calculation with both to satisfy yourself that you get the same answer). If the question tells you that one of the reagents was added in excess, then obviously don't use that number of moles.

Good luck, don’t panic, always think it through from first principles!