Same quantity of electricity is passed separately through a solution of a chloride of metal M and acidulated water when 14.8 litre H2 at NTP and 42 gram of M are obtained .if the specific heat of metal is 0.098 then determine the formula of mettalic chloride ?
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Reduction at cathode: 2H
+
(aq)+2e
−
→H
2
(g)
At N.T.P, 22.4 L (or 22400 mL) of H
2
= 1 mole of H
2
112 mL of H
2
=
22400
112
×1=0.005 mole of H
2
Moles of H
2
produced =
96500(C/mole
−
)
I(A)×t(s)
× mole ratio
0.005mol=
96500(C/mole
−
)
I(A)×965s
×
2 mol e
−
1 mol H
2
I=1A
Reduction at cathode: 2H
+
(aq)+2e
−
→H
2
(g)
At N.T.P, 22.4 L (or 22400 mL) of H
2
= 1 mole of H
2
112 mL of H
2
=
22400
112
×1=0.005 mole of H
2
Moles of H
2
produced =
96500(C/mole
−
)
I(A)×t(s)
× mole ratio
0.005mol=
96500(C/mole
−
)
I(A)×965s
×
2 mol e
−
1 mol H
2
I=1A
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