Recent content by Physiona

The molar mass of Na2CO3 is 106.. Is my method right, or have I done it differently/wrong?

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##

Thank you! That does make sense for that section. Just not entirely sure if my sine rule method is right. Am I going in the right lines?

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.…

I don't know the value of ##\theta## though.. Is the sine rule right..?

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...?

Why is it negative though? Does it always have to be under these circumstances?

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}##?

The sine rule in which I've been taught (using the symbols you've given now): ##a##\##\alpha## = ##b##\##\beta## = ##c##\##\gamma##..?

##x/\sqrt3/2## = ##y##/Sin(Theta)? Or if I'm intending to use, the other half: ##3x/\sqrt3/2## = ##x/\sqrt3/2##?