- Messages
- 160
- Reaction score
- 197
- Points
- 38
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All the A 2's please help me in this questionView attachment 39677
Can someone describe the reactions in periodicity and group 2 metals with oxygen, water and chlorine? And the observations
Dont u think u have written about Gp 1Group 2
Reaction with Oxygen:
4Na2 + 02 -----> 2Na20 (Yellow flame)
* Turnish in air
Burn on heating with Yellow flame
Form a ionic oxide ( Sodium oxide is a white solid)
Reation with chlorine
2Na2 + Cl2 -----. 2NaCl
*react on heating (white solid sodium chloride form)
Form ionic Chloride
2)Reation with water:
Na2 + Cl2 -----> 2NaOH + H2
When a piece of sodium is place on the surface of water it floats and fizzes about. It burn with an oragne flame.
Reaction with others i will soon posted if i had time. Got to go..
I misread it . I hope not to do in CIE exam
What about the color of the flames? Only magnesium is yellow and sulfur is blue.. Do the rest burn with a white flame?In Gp 2 the reactivity and reducing power increases down the group melting n boiling point decrease due to strength of metallic bonding
have low densities compared to other metals bt higher than Gp1
Reaction with water or steam:
1) reactions are exothermic
H2 gas is produced
Mgo reaction with water is very slow when heated with steam Mgo produced which is white powder
btw metal +water always produce metal hydrooxide n hydrogen gas while metal +steam always produce metal oxide n H2
Gp2 reactions form weak alkaline solutions which are slightly soluble
Reaction with oxygen:
GP 2 ELEMENTS REACT VIGOROUSLYwith oxygen and forms oxides and all of thm are white
reaction becomes vigorous down the group
thermal stability:
Thermal stability of Gp2 elements' nitrates n carbonates increase down group
bond breaking og Mgo is easiest
down the group size increases polarising power increases reactivity and stability increases
thats all I know sorry dunno about cloride
the metaloxides formed can react with water to form respective hydro oxides
Could you still post them?Dont u think u have written about Gp 1
Idk both I guessYou're asking abt the period or group :/
Lol lemme check my notesIdk both I guess
Idk both I guess
i was just going to ask the same doubt but saw it being explained alreadyAaah. Just did this 5 mins ago
Okay so basically, first things first, you gotta realize that 3 water molecules will be lost, so the Mr of the fragment[gly-ala-ser since equal amounts] will become (75+89+105-(18*3)) = 215.
Dividing 600,000 by 215 will give you the number of fragments present. I'm carefully choosing my words here. Fragments. Not residues. Each of these fragments will have 3 residues so the total fragments present * 3 = number of residues.
Mathematically,
600,000/215 = 2791 fragments
2791 fragments = 2791*3 residues = 8373 residues.
I'm confused about where the C comes in the 2nd aldehyde (CX).CH3(CH2)7CHO
OHC(CH2)7CX
All you have to do is split the double bond and partially oxidise it i.e keep the end point as the Aldehyde.
1b:
If I'm not mistaken,
Ethanal: A
Ethanol: C
Methoxymethane: A
2-methylpropane: B
Not sure though. If I'm wrong, do let me know.
1d:
I'd probably go along the lines of Hydrogen bonding.
I'd write that ethoxyethane cannot form hydrogen bonds with water so two insoluble layers are formed.
That'd get me one mark I think. Not sure what the second mark is for.
5e:
So you basically need a di-ol.
ethanedial is CHOCHO so COHCOH pretty much. The displayed structure would have an alkyne with 2 OH R Groups
OH-C=-C-OH
=- <= triple bond.
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