Asked by shearlene
George needs 50 liters of a solution that has a concentration of 30 g/ml for the manufacture of computer parts. George has an unlimited supply of a solution with a concentration of 39 g/ml. Using the Formula C1*V1=C2*V2 answer the following questions. Round answers to the nearest 10th. Use Unit Abbreviation Sheet for Units
How many liters of 39 g/ml equals the other mixture? How many liters of water must be added to this amount of the 39 g/ml solution to obtain the desired solution?
How many liters of 39 g/ml equals the other mixture? How many liters of water must be added to this amount of the 39 g/ml solution to obtain the desired solution?
Answers
Answered by
bobpursley
you are diluting it 39/30 times, or one part original, 9/30 part water
what is one part? 50/(39/30) or
1500/39=38.46 liters orig solution
water= 38.46*9/30=11.54 liters water.
Now for the other question, another way:
grams of stuff in 50 l: 50*30Kg/L
= 1500kg in 50 liters
grams of stuff in 38.46 liters of the original: 38.46*39kg/L=1499.94 rounded=1500g
Now round the two volumes to the nearest tenth
what is one part? 50/(39/30) or
1500/39=38.46 liters orig solution
water= 38.46*9/30=11.54 liters water.
Now for the other question, another way:
grams of stuff in 50 l: 50*30Kg/L
= 1500kg in 50 liters
grams of stuff in 38.46 liters of the original: 38.46*39kg/L=1499.94 rounded=1500g
Now round the two volumes to the nearest tenth
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