Explanation

(a)(i) ratio of atoms

C = \(\frac{0.48}{12}\) = 0.04, H= \(\frac{0.08}{1}\)  = 0.08, Cl = \(\frac{1.42}{35.5}\) = 0.04

Divide by the smallest number

\(\frac{0.04}{0.04}\): \(\frac{0.08}{0.04}\) : \(\frac{0.04}{0.04}\)

1 : 2 : 1

Empirical formula = CH\(_2\)Cl      

(ii)  Molecular formula = (Empirical formula)n
99 = (CH\(_2\)Cl)n     
99 = (12 + (2 x 1) + 35.5)n  
99 = 49.5n
n = \(\frac{99}{49.5}\)
= 2                       
molecular formular = C\(_2\)H\(_4\)Cl\(_2\) / (CH\(_2\)Cl)\(_2\)     
                                                                                                           

(b)       

-           in its molten form / solution conducts electricity
-           has strong electrostatic forces
-           has high melting / boiling points
-           is highly soluble in water

(c)        (i)  & (ii)

Reaction co-ordinate
                                   
                                               
(d)        (i)        Naphtha Accept kerosene / Diesel        
            (ii)        Fractional distillation               
            (iii)       -           Ethene            
                        -           Propene          

(iv) Catalytic cracking whereby long-chain hydrocarbons (in the naphtha fraction) are broken down into small organic molecules
Conditions are 600\(^o\) C and Al\(_2\)O\(_2\)\(_{(s)}\) / SiO\(_2\)\(_{(s)}\) as catalyst

    
(v)       

-           Polyethene
-           Polypropane

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