A pharmacist has 30% and 60% iodine solutions on hand. How many liters of each iodine solution will be required to produce 6 lit
ers of a 50% iodine.
1 answer:
Answer:
4 liters of 60% solution; 2 liters of 30% solution
Step-by-step explanation:
I like to use a simple, but effective, tool for most mixture problems. It is a kind of "X" diagram as in the attachment.
The ratios of solution concentrations are 3:6:5, so I've used those numbers in the diagram. The constituent solutions are on the left; the desired mixture is in the middle, and the numbers on the other legs of the X are the differences along the diagonals: 6 - 5 = 1; 5 - 3 = 2. This tells you the ratio of 60% solution to 30% solution is 2 : 1.
These ratio units (2, 1) add to 3. We want 6 liters of mixture, so we need to multiply these ratio units by 2 liters to get the amounts of constituents needed. The result is 4 liters of 60% solution and 2 liters of 30% solution.
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If you're writing equations, it often works well to let the variable represent the quantity of the greatest contributor—the 60% solution. Let the volume of that (in liters) be represented by v. Then the total volume of iodine in the mixture is ...
... 0.60·v + 0.30·(6 -v) = 0.50·6
... 0.30v = 0.20·6 . . . . subtract 0.30·6, collect terms
... v = 6·(0.20/0.30) = 4 . . . . divide by the coefficient of v
4 liters of 60% solution are needed. The other 2 liters are 30% solution.
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Answer:
The volume of the solid is 243
Step-by-step explanation:
From the information given:
BY applying sphere coordinates:
0 ≤ x² + y² + z² ≤ 81
0 ≤ ρ² ≤ 81
0 ≤ ρ ≤ 9
The intersection that takes place in the sphere and the cone is:
Thus; the region bounded is: 0 ≤ θ ≤ 2π
This implies that:
ρcosФ = ρsinФ
tanФ = 1
Ф = π/4
Similarly; in the X-Y plane;
z = 0
ρcosФ = 0
cosФ = 0
Ф = π/2
So here;
Thus, volume:
V = 243