Page 277 - Chemistry

CT – 2
D – 1
A – 1
T – 1
S = ½
P = 1
C = 1

(ii) 1 = 0.00510 √½ From the graph and must be shown. Showing. √½
t

t = 1 √½ = 196.5 seconds. √½
0.00510
(iii) Mg (s) + H 2SO 4(aq) MgSO 4(s) + H 2(g) √½
1 : 1 With correct physical state.
(iv) Moles of Mg = 0.12 √½ = 0.005 moles √½
24 1mk
Moles of H 2SO 4 used = 0.005 moles (1 : 1)
(v) Increase in length of M of ribbon results in decrease in 1
t √½

This is done to gradual decrease in the concentration of the acid. √½

Table II

Titration I II III
3
Find burette reading (cm ) 15.3 30.5 45.7
Initial burette reading 0.0 15.3 30.5
3
Volume of solution B used (cm ) 15.3 15.2 15.2
CT = 1
D = 1
AC = 1
PA = 1
TA = 1
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