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Summary CCEA GCE Chemistry AS module 1 (physical & inorganic chem.) (units 1-6) $7.53   Add to cart

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Summary CCEA GCE Chemistry AS module 1 (physical & inorganic chem.) (units 1-6)

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These notes summarise the full CCEA GCE Chemistry AS module 1 (Physical and Inorganic Chemistry) course. I wrote and utilised them to revise for my Chemistry AS exams (taken in the summer of 2022) in which I received an ‘A’ grade. The notes are handwritten using GoodNote and contain diagrams. I...

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  • October 28, 2022
  • October 28, 2022
  • 16
  • 2021/2022
  • Summary
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Chemistry U1
&




1.1 - Formulae, equations & amounts
1.2 - Atomic Structure





1.3 - Bonding
1.4 - Intermolecular Forces
1.5 - Structure
1.6 - Shapes of Molecules




Patrick Davis, 2022

, Unit 11: formulae, equestions and amounts.

-
Calculations
stones:
Calculating no.
of ©
6.02x10-23
No,
particles (or stones) urls
· = =
of
6.02x1023 costero
arogedo's
· =




male: not
use, the
partroctors/product
note - in

coles:
1) Write baloned
eye. Example: Conversion otoner
rent to iron

2). Cobulate no 13. Jezob(s> +bConeg)
greatly t320g) [faxes
->



demical be 2) Moles fez0z = wile= Pr =

16x108 - 168 =
100,000 md.
3). Write strin 3) Inc fezOz
egg. storian: = Inalfe
4) Convert need them ·
200,000 nd f2z0z =
200,000 rel Fe.
of
t mass 4). Massfe: res +Mr =2x105 x 56= 112 tonnes fe
y
postouts cales:
Limiting
1) Write balanced eye. Example: 100foy Huenite +
T0toL= From +CO2
2) Corulate
redesg ALL 1).2fez0y(s) +34s> > 4 fecs) +3C02 (y)
portant 2) Mdefezoz =
200x10g=(2250) +(3x= 4.3TNPTrol
3) x108md.
Identify lity retai roles L
T0x106y=12 5,836
= =


of
i) Write stori 3) Storid: 2nd
FezOz = 3nd(
Cyn
ii). Divide actual arles
of
actual: O' X0
rectants
by smalles 4-37x10S 4.33x 106
value A
get 1:1 patio =InofezOz 133rd (
=
a



iii) Compare store IndEazOz 266mrC:C:linity
= locat
=>
ratio its

4) Stoi: CE4ndfe
octer ratio
bidetjy brd

the
hinting rectors = 5.833 x18mC=(8bb) x 4 nel fe

4) Here relevant stain ration to : there are 7.778x0olfe.
uses unknown 5) Masxfa= TTT8x00
x56=4356x108g
find of
5. Comert rulesto woes.
= 436 tonnes 2.
of
Calculating males and Mo


Mass Mr= odd all the roll
Moles= of
Ma numbers
together



Patrick Davis, 2022

, Water
of Crystaliation 1


hydrate comproads:
Common

(I) [uSOy. 5H2O blue
Hycraled
=


eyote
-
·
=

copper
Hyedel zobalt (I) corile <0C.6Hz0 pint
·
-
= =


©

Hydrated sodium Carbonate: NrzCOy. 10H20:cdourless
·




·



Anhydrous:
Contois no
Hz0 of crystalisation
does contain
HD of hydration.
Hyerated:
·




·


Experiment to
find mass
of H2O o caytellication
17. enable tie.
Weight empty
+




EAAforofar
as
2)Add
5g of Hydrated InSOx +reneigh
3) set
diagram
the
up opportes us in
port
heat cresible over a rechine
glove (sitch t
moisture)
rooming later, A drive
of busen-Heat -latproof
4) Allow cool
after 15 mins + reeigh
5). Heat until constant moss
ogin for permis a+
pasigle-repeat


finding formulae hydrated compounds:
·



of
·

Example: Hydhotel Copper() ebote crystals contain 63.9%. GSO4
and
36.1720. Calculate is
empirical formula.
·


Compound: Copper (II) subate Water
·

formula: CuSO4 HeO
·
Mass: 63.9
63.9g. 36.1 %
36-18
% = =

·

Mr: 64 +32 + (16 x4) = 160 18
·

Moles: 63.9 = 160 361=18
=
0.399 rel. t =
200G nel.

· whole no 0.39 9 0.399
ratio: = I rel to =5 ne
·
Rio
of
compounds: Ina CuSO4 to 5 rel
H2O
Empirical
·




formule: CuSOp. 5H, O




Patrick Davis, 2022

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