Asked by Alex
Sulfuric acid is diluted in water to make new mixtures. In one container, 72 ml of a 14% acid solution and, in another container, 64 ml of a 31$ acid solution. The contents of the containers are mixed together? What's the percentage strength of the acid in the mixture?
Answers
Answered by
Bosnian
14 % = 0.14
31 % = 0,31
Total volume of containers:
72 ml + 64 ml = 136 ml
Volume of acid in first container:
72 ml * 0.14 = 10.08 ml
Volume of acid in second container:
64 ml * 0.31 = 19.84 ml
Volume of alcohol in the mixture :
10.08 ml + 19.84 ml = 29.92 ml
Percentage strength of the acid in the mixture:
( 29.92 / 136 ) * 100 % = 0.22 *100 % = 22 %
31 % = 0,31
Total volume of containers:
72 ml + 64 ml = 136 ml
Volume of acid in first container:
72 ml * 0.14 = 10.08 ml
Volume of acid in second container:
64 ml * 0.31 = 19.84 ml
Volume of alcohol in the mixture :
10.08 ml + 19.84 ml = 29.92 ml
Percentage strength of the acid in the mixture:
( 29.92 / 136 ) * 100 % = 0.22 *100 % = 22 %
Answered by
scott
.14 * 72 = ? mL acid
.31 * 64 = ? mL acid
64 + 72 = ? mL mixture
divide the TOTAL mL of acid by the mixture volume to find the % acid
.31 * 64 = ? mL acid
64 + 72 = ? mL mixture
divide the TOTAL mL of acid by the mixture volume to find the % acid
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