Asked by Jake
To dilute a solution from 0.600 M to 0.100 M, the final volume must be
A. 1/6 the original volume.
B. 0.7 L.
C. More information is needed to answer this question.
D. 6 times the original volume.
E. 60 L.
A. 1/6 the original volume.
B. 0.7 L.
C. More information is needed to answer this question.
D. 6 times the original volume.
E. 60 L.
Answers
Answered by
DrBob222
Just pick a number like 100 mL
mL x M = mL x M
100 x 0.6 = mL x 0.1
Solve for mL and I get
100*0.6/0.1 = 600
600 mL is what times 100?
mL x M = mL x M
100 x 0.6 = mL x 0.1
Solve for mL and I get
100*0.6/0.1 = 600
600 mL is what times 100?
Answered by
Anonymous
d
Answered by
Anonymous
The concentration of Pb2+ in a sample of wastewater is to be determined by using gravimetric analysis. To a 100.0-mL sample of the wastewater is added an excess of sodium carbonate, forming the insoluble lead (II) carbonate (267 g/mol) according to the balanced equation given below. The solid lead (II) carbonate is dried, and its mass is measured to be 0.1443 g. What was the concentration of Pb2+ in the original wastewater sample?
Pb2+(aq) + Na2CO3(aq) → PbCO3(s) + 2Na+(aq)
Group of answer choices
Pb2+(aq) + Na2CO3(aq) → PbCO3(s) + 2Na+(aq)
Group of answer choices
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