chemistry

[nid404]

with cold KMNO₄ it'll form a diol

CH₂OH CH(OH) CH₃

with bromide cylcohexene's double bond breaks and Br joins to one of the carbons of the bond.



with hot KMNO₄

bond breaks, forming dicarboxylic acid....

[tmisterr]

b(ii) is aluminium oxide
b(iii) is sulphur trioxide (it forms sulphuric acid).....it cant be any of the others because the first two are basic, Aluminium oxide is insoluble and Sulphur dioxide form sulphurous acid
c(ii) its an acid base reaction to form a salt and water
        SO₂ + NaOH ---> NaHSO₃ + H₂O (Unbalanced)
d i 9O₂ and 6SO₂ to balance
  ii 4x+(6*-2)=0 solve for x, x=+3
  iii Work with mole ratios here.
    10 moles of Sb₂S₃ will produce 10/2=5 moles of Sb₄O₆, carry down the 5 moles to the next equation and five mole of Sb₄O₆ will produce 5*3= 15 moles of Carbon dioxide. at rtp one mole of a gas occupies 24 dm³ so 15*24 will be 360dm³ of carbon dioxide

[ruby92]

b(ii) is aluminium oxide
b(iii) is sulphur trioxide (it forms sulphuric acid).....it cant be any of the others because the first two are basic, Aluminium oxide is insoluble and Sulphur dioxide form sulphurous acid
c(ii) its an acid base reaction to form a salt and water
        SO₂ + NaOH ---> NaHSO₃ + H₂O (Unbalanced)
d i 9O₂ and 6SO₂ to balance
  ii 4x+(6*-2)=0 solve for x, x=+3
  iii Work with mole ratios here.
    10 moles of Sb₂S₃ will produce 10/2=5 moles of Sb₄O₆, carry down the 5 moles to the next equation and five mole of Sb₄O₆ will produce 5*3= 15 moles of Carbon dioxide. at rtp one mole of a gas occupies 24 dm³ so 15*24 will be 360dm³ of carbon dioxide

im sorry i attached the wrong paper its actually october/nov 2003 
here's the paper

[ruby92]

m/j 2004 paper 1
Q23
26
28
35
39 (why only the first one why not the second one; theyre both aldehydes)

[nid404]

m/j 2004 paper 1
Q23
26
28
35
39 (why only the first one why not the second one; theyre both aldehydes)

23) straight D
Cracking of a nonane can nvr give u a hydrocarbon with 10 carbon atoms...
26)straight D again....
slower reaction=high Ea
faster reaction=low Ea
28) D agn
monosubstituted alkene, one R group is attached to the carbon-carbon double bond. Not oxidized by mild oxidising agents
35) B
The temperature of the process causes the decomposition of the calcium carbonate
into calcium oxide.
Hence not 3
39) D straight
When you see a benzene ring, u knw it isn't soluble in water. So, 2 and 3 are out

[tmisterr]

b ii Q=m*c*change in temp so 12.2*4.2*200=10248 j
b iii 10.248kj is the enthalpy change when 1 g of calcium is dissolved, but one mole of calcium is 40.1g  so 40.1*10.248=410.9 = 411 kj mol^-1
c ii you need to use a hess cycle
      We know the enthalpy change of the reaction to form Calcium Hydroxide and Hydrogen (calculated in biii) and we have been given the enthalpy change of formation of water.
                       
                               Ca + 2H₂O ----> Ca(OH)₂ + H₂

                                                     
                                              Ca + 2H₂ + O₂

so applying Hess law, starting from the lower equation, you can either first form calcium and water then form calcium hydroxide or directly form Calcium Hydroxide(this is the enthalpy change we are looking for). therefore:
                    2*-286+-411= enthalpy change of formation of Calcium hydroxide = -983kj mol^-1

d) mole ratios (from equation in b)
 1 mole of calcium forms one mole of hydrogen so 1/40.1 moles of calcium will form 1/40.1 moles of hydrogen.
so volume of oxygen is 1/40.1 * 24= 0.5985 dm³ or 598.5 cm³

[ruby92]

28) D agn
monosubstituted alkene, one R group is attached to the carbon-carbon double bond. Not oxidized by mild oxidising agents

could you explain this a bot further please?

[chanella]

http://www.a-levelchemistry.co.uk/AQA%20AS%20Chemistry/Unit%201/1.5%20Introduction%20to%20Organic%20Chemistry/1.5%20home.htm

This website has notes,worksheets, tests

[ruby92]

Thank You ppl 

[ruby92]

o/n 2008 paper 1
4 9 13 19 39(how is 2 correct) 32 ( water acts as a base rite?) 29 (why not A), 26(why not B)

[nid404]

Q4) C

H has 1
C has 6
O has 8 (times 3)

and the overall charge is -1 (which means there's one more electron)

1+6+24+1=32

Q9)  A

just the reverse right....so greater activation energy (endothermic reaction) and so

Q13) Phosphoric acid is more acidic as compared to the others.  D

Q19) ammonium ion acts as an acid ( proton donor) H+ ion donor. hence D

Q39) 2nd one is an isomer of the first...so it's absolutely ok.

Q32) water is amphoteric. it can act as an acid and also a base. It can donate H+ ions and can also accept them forming(hydronium ion)

Q26) count the no of Carbon atoms in the product formed...lesser than the one they are asking for

Q29) again count the no of carbon atoms. In A the C from CN is not present. The count of C atoms should increase by 1 due to addtion of CN- ion

[ruby92]

no 2 could you explain it in a bit more detail.
m/j 2008
34
39(why is 3 not correct?)
38
29 why not c
21 15 9

[nid404]

which one did you not understand?

[ruby92]

sorry 26

o/n 2007
4 isnt it D group 3
5
9
37

[nid404]

the product they specified has 1 more carbon atom than the molecule in B