nope its grammar 2020is this the SGS trial haha?
WAIT IS SGS GRAMMAR??? LMAO IM STUPID ASnope its grammar 2020
ohhhh okay that makes sense - thanksAh found it, so i got quite stuck on this too, so basically adding 10ml of solvent x dilutes the concentration of I2 in the solution. So to increase the Concentration again, the Eq shifts to favour the forward reaction, decreasing the conc of I2 in Y
Is there a lone pair at 3 by any chance?
I didn't think so, because if you look at the number of bonds to the C, it has four (the C=O, CH-C, C-OH).Is there a lone pair at 3 by any chance?
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also this <3
so trueI didn't think so, because if you look at the number of bonds to the C, it has four (the C=O, CH-C, C-OH).
Yeah so I said Ba(OH)2 + NaOH --> Na+ + Ba2+ + 3(OH-) --but had no clue what to do from thereSorry it's been a while since I did this paper lol, I *THINK* This is because NaOH produces 1 hydroxide ion when dissolved in water and Ba(OH)2 produces 2 so they just add up
Yeah so I said Ba(OH)2 + NaOH --> Na+ + Ba2+ + 3(OH-) --but had no clue what to do from there
hmm no I still don't really get it: I have 1:1:3 of Ba:Na:OH, don't I?I think it wouldn't add up to 3(OH) per say but just the total concentration of OH ions in the solution if that makes sense
still don't get it :/ sorrywell yes, but the question is asking for the "concentrations of each of the ions in the solutions formed" so its not asking for the ratio but the final conc?
no you're good dw, if you let me think on how to better explain it i might get back to you tn, unfortunately i do have to cram the mod 8 ions flowchart into my head and cry about ext 2 mathsstill don't get it :/ sorry
also i'm def sure its dilutealsooo is hydration of alkene to alcohol with conc or dilute H2SO4??