I remember like this, oxidising agents are elements like OXYgen. Oxygen needs only 2 electrons to gain full shell stability so since it gains it gets reduced. If it gets reduced it oxidises others therefore its an oxidising agent
Hey, wondered if anyone could clear this up for me,
Going over past papers for Chem6x I've come across drawing graphs for the Iodine clock reaction a few times, usually pretty straightforward, log (1/t) is plotted on the y-axis and Log(volume) on the x-axis, but I noticed in the mark scheme that says a mark is lost if the y-axis is plotted with ascending negative numbers? I'm not sure if the wording is awkward but the y-values for Log91/t) are usually negative, it doesn't seem to make sense to plot them in a descending scale (i.e. negative values getting more negative going up the y-axis),
Cheers
I think that you have to plot the graph with the axis at the top if you see what i mean, so like a normal axis but upside down and then put your negative values on the y axis getting more negative as you go down
Well the ideal solvent is the one in which solute dissolves most in hot solution and doesn't dissolve in cold solution at all. Since no solution is ideal, we use the solution in which most solution dissolves most in hot solution and least in cold solution, as in the one with greatest difference in the two solubility values. So now in this case, the difference in values of solution 1 is 5 whereas in solution 2 is 7.5 and as solution 2's difference > solution 1's difference so 2 is a better solvent. Well i hope it was a helpful explanation
Can someone help me with q9? It's about percentage error, and they've said in the answer that the titre will be 24.4 so you can use that. I've forgotten what the formulae for percentage error is :// http://filestore.aqa.org.uk/subjects/AQA-CHM6T-P10-TEST.PDF
Can someone help me with q9? It's about percentage error, and they've said in the answer that the titre will be 24.4 so you can use that. I've forgotten what the formulae for percentage error is :// http://filestore.aqa.org.uk/subjects/AQA-CHM6T-P10-TEST.PDF
Thanks
Well percentage error is (uncertainity or error)/total titre*100 and since this gives percentage error in each reading so when doing this for burrette we multiply uncertainity by 2 as we take two readings (initial and final)
You should definately watch a few videos on this if you haven't done it in labortary yourself. One i'd suggest is https://youtu.be/qJLvB6NFnoA And you'll see that to dissolve the solute we heat the solvent x)