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Chemistry: Post your doubts here!

Messages
675
Reaction score
862
Points
103
All the A 2's please help me in this questionView attachment 39677

Simple! It's pretty much an AS question.

E is a di-ol. Just add 2 OH's across the double bond.

ii)
Reaction I: Cold + dilute KMnO4
Reaction II: K2Cr2O7 + H+ + warm


__

Incase you don't get whats going on, the cold KMnO4 will add a di-ol across the double bond. Cr2O72- + heat oxidises primary alcohols to aldehydes (if reflux then carboxylic), secondary to ketones, and tertiary to nothing. The upper ol is primary and the lower is tertiary so stays as is.
 
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872
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Can someone describe the reactions in periodicity and group 2 metals with oxygen, water and chlorine? And the observations
 
Messages
2,222
Reaction score
4,914
Points
273
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 :oops:
the metaloxides formed can react with water to form respective hydro oxides
Can someone describe the reactions in periodicity and group 2 metals with oxygen, water and chlorine? And the observations
 
Messages
2,222
Reaction score
4,914
Points
273
Group 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..
Dont u think u have written about Gp 1
 
Messages
872
Reaction score
894
Points
103
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 :oops:
the metaloxides formed can react with water to form respective hydro oxides
What about the color of the flames? Only magnesium is yellow and sulfur is blue.. Do the rest burn with a white flame?
 
Messages
2,222
Reaction score
4,914
Points
273
I think you're asking abt period 3 :/
REACTION OF PERIOD 3 WITH OXYGEN
sodium reacts vigorously and burns in a yellow flame and white solid formed

Mg burns in a dazzling white flame and white solid formed

Al is protected by an oxide layer
but powdered Al burns with a bright white flame

Si reacts very slowly n forms grey solid

Phosphorus has 2 allotropes white phosphorus is very reactive while red P is less reactive

white phosphorus burns with a yellow flame cloud of white phosphorus(V) oxide formed

sulfur burns gently with a blue flame in gas jar of O2
this reaction takes place in precense of catalyst at 450°c

Chlorine n ardon dont react with O2
hope it helped :/
Idk both I guess
 
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253
Reaction score
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Aaah. Just did this 5 mins ago :D

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 was just going to ask the same doubt but saw it being explained already :D
 
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119
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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.
I'm confused about where the C comes in the 2nd aldehyde (CX).

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.

When do we know if it is a permanant or induced dipole?

Yeah, I got that. Problem is the second mark. In the marking scheme it says " there is hydrogen bonding between H2O molecules" but I don't get that.

Thank you, I got confused for the triple bond. So we only have to derive the bond basically.
 
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9701/12/O/N/10
Question 3
 

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