a 5.00 ml sample NAOH was neutralised by 20.52 ml of a 0.952 m L-1 HCL.
find conc. of NAOH.
i worked it out as following:
n(HCL) = c*v
= 0.952 * 0.02052
=0.0195
n(NAOH)= n(HCL)
C(NAOH) = n/v
= 0.0195/0.00500
=3.9 mol L-1
Is that 3.9 possible ...
i didnt no u had to know how to make it- like set it up. Thats a bummer.
Also im guessing their gonna ask a question based on the differences of electrophoresis and chromatography
we did the simulation too... so maybe you just write about using restriction enzymes to make them smaller pieces and then putting them in the gel to be seperated ... and then talk about why they are seperated
maybe...
i dont care about them asking for a uai estimation/ prediction, but whats the point in asking " if i study heaps hard will i go well" ?
Of course if u study heaps hard u will go well