chem Flashcards
(149 cards)
emission spectrum of hydrogen
coloured lines of discrete frequencies/energy levels in a dark background, which correspond to electron transition from higher to lower energy levels.
Coloured lines converge at higher frequencies as energy levels are close together at higher energy levels
lyman
emission to n = 1
balmer
emission to n = 2 (visible light)
paschen
emission to n = 3
oxides and water
acidic oxide = acid
basic oxide = base
amphoteric oxide = insoluble, reacts with both acid and base BALZ barium aluminium lead zinc
whenever you see a find Kc qns
ICE table damnit (values are in concentration)
Gibbs free energy unit
kJmol-1 (around 3 digit, some formula gives u in Jmol-1)
how to find conc. of iodine with formula
titration with sodium thiosulfate
I2 + 2S2O3(2-) -> 2I(1-) + S4O6(2-)
aiya see onenote search ‘reaction of iodine with thiosulfate
standard electrode potential
standard electrode potential of a half-cell is the potential difference generated when it is connected to the standard hydrogen electrode at standard conditions
finding pH or H+
14 - pOH
[H+] = sqrt(Ka x (C-x))
Kw = [H+][OH-]
(polar) Protic solvent favoured by
solvates nucleophile and carbocation intermediate Favours sn1 (reduces Ea of rds cause carbocation is more stable) Reduces effectiveness of sn2 (nucleophile gets solvated)
(polar) Aprotic solvent favoured by
solvates cation counterion favours sn2 (increase reactivity of nucleophile to donate a pair of electrons)
transition metal r/s with periodic table group
group no. is no. of electrons in 3d + 4s orbitals
Electrophilic Addition: why is there major product
tertiary>secondary>primary (carbocation intermediate)
__ carbocation is more stable due to positive inductive effect caused by more electron-donating alkyl groups that disperse the positive charge on carbocation
reagents and conditions: nucleophilic substitution
NaOH, heat under reflux
KOH, heat under reflux
rmb to add lone pair and negative charge on transition state for mechanism
why atoms can exceed octet rule
period 3 and onwards (presence of 3d orbitals)
strong acid + weak base =
acidic salt (strong conjugate acid, POE to left, will react with OH to return back into a weak base, reduce pH)
salt hydrolysis: salt + h2o will give H3O+ thats why acidic
weak acid + strong base =
basic salt (strong conjugate base, POE to left, will react with H to return back into a weak acid, increase pH)
salt hydrolysis: salt + h2o will give original acid n OH- thats why basic
strong acid + strong base =
neutral salt
both have weak conjugate base/acid, so [H] : [OH] remains same
weak acid + weak base
neutral salt
both have strong conjudate base/acid, [H] : [OH] stay same
given Ka of weak acid, how to find Kb?
Kb is of the strong conjugate base
Ka = Kw/Kb
pKa = pKw - pKb
s. f of absolute uncertainty
* rmb to follow d.p of absolute uncertainty
1sf, round mathematically, dont always round up
(like in IA)
*rmb to follow d.p of absolute uncertainty
first order wrt a rxt
constant half-life
half-cell potential meaning (more +/- is…) and Ecell formula
more positive = reduction
more negative = oxidation
Ecell = reduction - oxidation
*rmb not to swap the sign, just plug into formula