Chemistry: Post your doubts here!

[techgeek]

Somehow I can't get any of the answers provided in question 5.

 

[Error Syntax]

here you go

 

[Amir Anwar]

what will be the product of HOCH2CH(OH)CO2H+CH3CO2H
m/j/09/21/q4 b2

 

[darks]

what will be the product of HOCH2CH(OH)CO2H+CH3CO2H
m/j/09/21/q4 b2

CH3COOCH2CH(OCOCH3)COOH ????

 

[Thought blocker]

CH3COOCH2CH(OCOCH3)COOH ????

Yes.

 

[student2]

I assume this is a simple concept. To know whether an atom donates a pair of electrons or shares them to expand it's octet, you have to think about each atom and how many electrons it requires to complete it's octet.
the atom that donates it's pair of electrons has to have at least one pair of lone pair of electrons after it has formed the bond.
also you have to know, that in a dative bond, the electron donor atom donates electrons in pairs. (in the case you mentioned, 3 electrons donated are not in pairs. hence in this case chlorine has to share those 3 electrons with 3 fluorine atoms to expand it's octet.) if an atom has those qualities metioned then in this case it will donate a lone pair instead of expanding it's octet.

you can look at other similar example to help you,

• like Al2Cl6, ( in this case there is no octet expansion taking place)

here each aluminium atom is covalently bonded with 3 chlorine atoms , while at the same time it's accepting a pair of electrons from a forth chlorine atom,,, take a look at the fourth chlorine atom, (as mentioned before the atom donating the pair of electrons has to have at least one pair of electrons after forming the bond, at the same time it donates the electrons in pairs.)
I hope this answers your question

z ths still dere in 2016 syllabus? i cnt fynd it in 2015 syllabus either...

 

[Holmes]

9701/12/O/N/10
October 2010 paper 1 : Q8 ,Q11 , Q28, Q34, Q39.
Help me please!!
Thanks in advance

 

[techgeek]

9701/12/O/N/10
October 2010 paper 1 : Q8 ,Q11 , Q28, Q34, Q39.
Help me please!!
Thanks in advance

Q8:
It's D. First IE of calcium = 590
Second IE of calcium = 1150

To convert solid Ca into gas you need 177 kJmol-1 energy, and then to convert it to 2+ ions you need 590+1150 energy. Then for hydrating the ions the energy change is 1565kjmol-1. So total energy change = 177+590+1150-1565 = +352 kJmol-1

Q11: Initial concentration of X: 0.50
concentration of X during equilibrium: 0.25
Initial concentration of Y : 0
concentration of Y during equilibrium: since 1 mol of X produces 2 mol so conc of Y= 0.25x2 = 0.50
Kc= [Y}^2/[X] =0.50^2/0.25 =1
So it's C


Q28: In skeletal formula, we don't display Hydrogen atoms, so A is incorrect.
C shows 5 carbon atoms while D shows 3, but the repeat unit shows there should be only 2 carbons with one florine which is in B only.

Q34: The reactions will be
Mg + 2 CH3COOH <=> Mg(CH3COO)2+ H2(g) (since ethanoid acid is a weak acid, it will be in equilibrium)
Mg + H2SO4 >> MgSO4 + H2
1 After 2 minutes, the sulfuric acid is at a higher temperature than the ethanoic acid.
2 After 2 minutes, the sulfuric acid has produced more gas than the ethanoic acid
Sulfuric acid is a stronger oxidizing agent and a diprotic acid while ethaonic acid is monoprotic, that means it will give off more hydrogen ions per unit time than ethanoic acid so more hydrogen gas will be produced. Since bond formation is exothermic, sulfuric acid will also be at a higher temperature
3.After 20 minutes, the sulfuric acid has produced more gas than the ethanoic acid.
The reaction between ethanoic acid and magnesium is in equilibrium, so it never reaches completion.
So the answer is A.

Q39: It's just calculation, find Mr for each of the reactants and products
no. of moles of reactants/no. of moles of products x 100%
you'll find they all give 62% yield so the answer is A.

that's it

 

[amina1300]

A space shuttle’s upward thrust came from the following reaction between aluminium and ammonium perchlorate.
10Al + 6NH4ClO4 → 4Al 2O3 + 2AlCl 3 + 12H2O + 3N2
Which statements about this reaction are correct?
1 Aluminium is oxidised. 2 Chlorine is reduced. 3 Nitrogen is oxidised.

 

[Thought blocker]

A space shuttle’s upward thrust came from the following reaction between aluminium and ammonium perchlorate.
10Al + 6NH4ClO4 → 4Al 2O3 + 2AlCl 3 + 12H2O + 3N2
Which statements about this reaction are correct?
1 Aluminium is oxidised. 2 Chlorine is reduced. 3 Nitrogen is oxidised.

 

[Holmes]

A space shuttle’s upward thrust came from the following reaction between aluminium and ammonium perchlorate.
10Al + 6NH4ClO4 → 4Al 2O3 + 2AlCl 3 + 12H2O + 3N2
Which statements about this reaction are correct?
1 Aluminium is oxidised. 2 Chlorine is reduced. 3 Nitrogen is oxidised.

Option D is only correct. i.e.
1 Aluminium is oxidised. (only)
why
If we look at the oxidation state of Al than we can clearly see that Al is oxidised from 0 to +3
while chlorine's oxidation state remains the same.i.e. -1
And Nitrogen is reduced from +5 to 0 while statement 3 says the opposite.
so answer is D

Hope it helped.

 

[Lee Qian Yi]

can explain why answer is C?

 

[Holmes]

can explain why answer is C?View attachment 61871

KMnO4 is a famous oxidising agent. Using COLD KMnO4 will break "C=C" and add an -OH group on each carbon atom. Now count the total number of hydroxy groups present. Two are attached with each Carbon and one already attached to your absolute LEFT hand side (in the diagram ).
So this narrows your choice and you are restricted to "C" and "D" only.
When hot ,concentrated acidified KMnO4 is used then this reagent will cause the "C=C" to break, thus making ketone. In this way ONE of the 6 - membered is broken. Now count the remaining 6-membered rings, and you will find that there are ONLY TWO (2) remaining 6- membered rings.
So your answer is "C".
I hope I helped you.

 

[Lee Qian Yi]

ok thx , i know already.

 

[Amir Anwar]

Yes.

how can u solve it

 

[Thought blocker]

how can u solve it

solve what?

 

[student2]

wht is da difference b/w H bonded alcohol nd free alcohol in IR spectroscopy? ny1 knws?

 

[Lee Qian Yi]

can explain why answer is D?

 

[abbas haider]

View attachment 61889
can explain why answer is D?

Read This






look at (ii) if if there is one alkyl group and one hydrogen at one end of the double bond a carboxylic acid will form ,now look at the question the organic compumd has one alkyl group and one hydrogen on either side of double bond this means two Carboxylic acids will form (COOH) howevver what this pic didnt tell is that if OH is already present in a compund it will always form a COOH.



Using elimination method
A) not possible because double bond has to rupture so 2 molecules have to form so this is not possible. (because its the only link between two group of atoms, THIS MAY NOT HAPPEN IN CYCLIC COMPUNDS)
B)double bond always rupture but this is nt the case since C=C is safe . so this is not possible.
C) although C is partially correct because if we have one alkyl group and one hydrogen at one end of the double bond , first aldehyde is formed , C=C breaks ,1 carbon forms COH. however the process oxidation hasnt ended . the -H in COH also gets an O forming -COOH . thus the end product is -COOH. thus c is not possible
D) Final answer and already told why.

 

[Lee Qian Yi]

oh

Read This

View attachment 61890




look at (ii) if if there is one alkyl group and one hydrogen at one end of the double bond a carboxylic acid will form ,now look at the question the organic compumd has one alkyl group and one hydrogen on either side of double bond this means two Carboxylic acids will form (COOH) howevver what this pic didnt tell is that if OH is already present in a compund it will always form a COOH.



Using elimination method
A) not possible because double bond has to rupture so 2 molecules have to form so this is not possible. (because its the only link between two group of atoms, THIS MAY NOT HAPPEN IN CYCLIC COMPUNDS)
B)double bond always rupture but this is nt the case since C=C is safe . so this is not possible.
C) although C is partially correct because if we have one alkyl group and one hydrogen at one end of the double bond , first aldehyde is formed , C=C breaks ,1 carbon forms COH. however the process oxidation hasnt ended . the -H in COH also gets an O forming -COOH . thus the end product is -COOH. thus c is not possible
D) Final answer and already told why.

oh i see , really thx very much !!