a sample of clay was partially dried and then contained 50 % silica and 10 % of water . the original clay contained 20 % water . find % of silica in original sample ( nearly ).
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answer ,
W ( clay ) + w ( water ) + w ( impurity ) = 100gm
w ( clay ) + w ( impurity ) = 100 - 20 = 80 gm.
upon heating only weight of water decreases .
now 80 g = weight of sample after heating × 90 / 100
weight of sample after heating = 80 × 100 / 90.
= 800 /9
weight of silica = 800 / 9 × 50 / 100
% of silica in original sample = 400 / 9
= 44%
be brainly
W ( clay ) + w ( water ) + w ( impurity ) = 100gm
w ( clay ) + w ( impurity ) = 100 - 20 = 80 gm.
upon heating only weight of water decreases .
now 80 g = weight of sample after heating × 90 / 100
weight of sample after heating = 80 × 100 / 90.
= 800 /9
weight of silica = 800 / 9 × 50 / 100
% of silica in original sample = 400 / 9
= 44%
be brainly
Answered by
13
Answer:
W ( clay ) + w ( water ) + w ( impurity ) = 100gm
w ( clay ) + w ( impurity ) = 100 - 20 = 80 gm.
upon heating only weight of water decreases .
now 80 g = weight of sample after heating × 90 / 100
weight of sample after heating = 80 × 100 / 90.
= 800 /9
weight of silica = 800 / 9 × 50 / 100
% of silica in original sample = 400 / 9
= 44%
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