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Chem paper 1 2005 doubts :D

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i need help on some questions :
From M/J 2005 : Q1, 2, 7, and 15
also from O/N 2005 : Q15, 16, 24 and 36. thanks guys
 
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Okay I can try to help you with the M.C.Q's one by one...for the M.C.Q 1 you should be aware of the fact that whenever a hydrocarbon is burnt the number of carbon atoms in that molecule are equal to the molecules of carbon dioxide produced, likewise the number of hydrogen atoms in the compound equal half the water molecules in the equation.

Using this rule, you can make equations for all the hydrocarbons listed quickly:
1. C2H6 + 3.5O2 ----------> 2CO2 + 3H2O
2. C3H6 + 4.5O2 ----------> 3CO2 + 3H2O
3. C3H8 + 5O2 ----------> 3CO2 + 4H2O
4.C4H10 + 6.5O2 ----------> 4CO2 + 5H2O

Hence now looking at the equations we can figure out the data mentioned in the question, the carbon dioxide produced is 30cm3 and the oxygen reacted had a volume of 50cm3 hence ratios are only fulfilled by equation 3 in which 1 mole of hydro carbon reacts with 5 moles of O2 to produce 3 moles of carbondioxide.

Hope it helps! :)
 
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For M.C.Q 2:

For this question just look at the equation mentioned in the question for once, if one mole of NaN3 is available then it would form one mole of Na and 1.5 moles of Nitrogen, then that same one mole of Na will be used in the next equation, hence if it is one mole then the nitrogen in the second equation would be 1/1O of Na which is 0.1 moles. Now add up the total moles, 1.5+0.1 = 1.6 and there you have it :)
 
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For M.C.Q 7:

Make the reaction route diagram firstly. I will try to make it here:

0.5 I2 + 1.5 Cl2 ----X----> ICL3

ICl + Cl2
X= -88 + 7
x= -81

We use seven since we need the energy per mole of ICl and the equation mentioned in question is for two moles. Hope it helps! :)
 
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For M.C.Q 15 May June 2005:

Concentrated sulphuric acid is powerful enough to oxidise HBr to Br2 and Iodine to I2 however it isnt powerful enough to oxidise HF and HCl. Its a fact :)
 
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For M.C.Q 36

Option 1 is correct since in NaClO3 the oxidation state of clorine is +5
Option 2 is also correct since the chlorine has two oxidation states in the products -1 and +5
Option 3 is incorrect since chlorine acts as the reducing agent since it oxidises itself.

Hope it helps! :)
 
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For M.C.Q 24:

I guess you just have to make an alkene out of each alcohol and make its isomers...three isomers are only possible in case A of butan-2-ol.
 
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M.C.Q 16:

For the limewater turning milky we need a carbonate and later in its remaining part a metal which produces hydrogen with the acid..I n which case the option is C with barium as a metal.
 
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For M.C.Q 15:

make the equation: 2Ca(NO3)2 ---------> 2CaO + 4NO2 + O2

the moles of calcium nitrate= 8.2/164 = 0.05
hence moles of oxygen= 0.025 moles
volume of oxygen= 0.025 * 24000 = 600 cm3

and youre done :)
 
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