Homework Statement
Sodium carbonate exists in hydrated form, Na2CO3.xH2O, in the solid state. 3.5 g of a sodium carbonate sample was dissolved in water and the volume made up to 250 cm3. 25.0 cm3 of this solution was titrated against 0.1 moldm-3 HCl and 24.5 cm3 of the acid were required...
Hello.
I'm assuming by the title of this thread that it isn't relevant to confess this certain topic here but I would like to get some advice.
I'm moving to sixth form and I have currently as my a level options Biology Economics and Statistics. I'm considering in changing these to Chemistry...
Right I got that so far.. As ##60+120=180##, final angle in the triangle is ##0##.
So we have ##\dfrac{\sin \theta}{x}=\dfrac{\sin (60°-\theta)}{3x}## (and the lengths and angles are the wrong way round in this formula of yours)
Do I cross multiply or simply rearrange in finding ##x##? I'm...
Sin##(60)## ---> is ##3x/sqrt 3/2## (EXACT VALUE) = \dfrac{y}{\sin 120}?##(Sin(60)## is similar to how others would approach it). By rearranging I get. ##sqrt3/2 y ## = ##sqrt 3/2 *3x##
Right, so ##\dfrac{y}{\sin 120}=\dfrac{3x}{\sin 60}=\dfrac{x}{\sin \theta}##, will become ##\dfrac{3x}{\sin 60}=\dfrac{x}{\sin \theta}##? Do I rearrange this to make ##\theta## the subject..? I'm confused as to how 60o has came from.
Aha riiiight, I get that working out now... (Yes i got confused between which factors I'm labelling it as..)
So now I have:
##\dfrac{y}{\sin 120}=\dfrac{3x}{\sin 60}=\dfrac{x}{\sin \theta}##
Is this correct? And can I ask how we obtained sin##(60)##?
Basically I've reached up to the sine rule I've made in post #10. Is that correct..? Under this case, I think I use the sine rule to find one length/angle. Not sure which one though.…
Aha right. That makes sense for that bit. Yes i apologise for the unclear diagram I did attempt my best to post a copy of the template; in the actual diagram the angle on the far right is C, angle of far bottom left is A and top left is B. Is my sine rule right so far...?
Yes sorry that's what I meant.
Hang on, I'm confused as to where ##3c## is from, didn't you label it as ##3x##? And ##\beta## isn't that sin##120## or something as from before?
##\dfrac{3x}{\sin 120}=\dfrac{x}{\sin 60}=\dfrac{y}{\sin \theta}##?