volume of oxygen at STP liberated by 5 ampere is flowing for 193 seconds to acidulated water will be
Answers
Answered by
6
I = 5A
T = 193 secs
Q = IT
= 5×193 = 965
Equivalent weight of oxygen = 8
Mass of oxygen liberated =
(E × Q)/96500g
= (8 × 965)/96500
= 0.08g
volume of 32g Oxygen at STP = 22400ml
Volume of 0.8 g oxygen at STP = (22400/32)×0.08
= 56 ml = 56 cc
Answered by
1
Answer:
56ml
Explanation:
no . of moles electrons =5 x 193÷96500 =0.01 moles
4 moles of electron = 1 mole of oxygen
volume of o2 at stp =22400ml
0.0025 moles =0.0025 x 22400 =56 ml
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