Please help!!! Chemical Mixture Problem!!!?

Question

If you are here to help me on this problem thanks! I need help on steps and the solution to this problem. Ive read the section it just doesnt help because the examples seem different PLEASE HELP!!!
HERES THE QUESTION:

The hospital pharmacist wanted 250ML of a soultion that was 72% iodine. How many milliters of a 40% solution should be mixed with how many Milliliters of an 80% solution to get the desired result?

THANKS IN ADVANCE! YOUR HELP MEANS ALOT!

Answer 1

200 ml of 80% gives you 160 parts of Iodine
50 ml of 40% gives you 20 parts of Iodine

200 + 50 ml = 250: you have the correct amount

160+20 = 180

180/250 = 72% you have your concentration.

I just felt that 200 ml and 50 ml were right, so I checked it with those numbers and sure enough, I got it!

Let's see if I can use algebra to work with it...

40% sol'n + 80% sol'n = 250 ml
Let x = the amount of liquid for 40% sol'n
x+80% sol'n = 250
therefore:
80% sol'n = 250-x

0.4x +0.8(250-x)=0.72(250)
0.4x +200 - 0.8x =180
-0.4x = -20
x= 50

The amount of 40% is 50 ml
80% amount is 250-50 = 200 ml

Answer 2

It's probably more useful to know how to work out the answer rather than just guessing...

Let's say we mix X mL of 40% iodine with Y mL of 80% iodine.

The total amount of solution is X + Y mL, which has to equal 250 mL.
The total amount of iodine is 40/100 X + 80/100 Y and this has to equal 72/100 * 250.

So, X + Y = 250
2/5 X + 4/5 Y = 180
Multiply the second equation through by 5 to get
2X + 4Y = 900
Now we know Y = 250 - X from the first equation, so we get
2X + 4(250 - X) = 900
=> 2X + 1000 - 4X = 900
=> -2X = -100
=> X = 50, Y = 250 - X = 200.
So we need 50 mL of 40% iodine and 200 mL of 80% iodine.

Answer 3

.4*x+.8*y=.72*250 and x+y=250 because the first equation is solving for how much iodine is in each and the second is for how much total liquid. From the second equation, you can get y=250-x and by substitution .4*x + .8*(250-x) =180 => .4*x + 200 -.8x = 180 => -.4x=-20 => x=40 and y=250-50=200 so you need 50 mills of the 40% solution and 200 mills of the 80% solution.

Answer 4

First, define your variables: Let's use F for the ml of 40% and E for the ml of 80%.

The strategy is to write an equation of THE AMOUNT OF IODINE, that is, the stuff in solution.

The amount of iodine in the 40% solution is .4*F (ie 40% of F ml).
The amount of iodine in the 80% solution is .8*E (ie 80% of E ml).
The total amount of iodine is in the final 72% solution: .72*250

The equation is the sum of the iodine in each soln adds to the total:

.4F + .8E = .72*250

That's one equation in 2 unknowns, so you need the second equation relating F and E. But that is straight forward as F + E = 250 (ie, the total amount of solution)

You can solve these two equations simultaneously, right?

Answer 5

i assume you dont could desire to look up NACL and parent atomic weight or no longer something this is unquestionably straightforward i think of in case you like 15% from 2 suggestions one 25 and one 5 nicely you're saying 15 is the avg of the two ie its the mid ingredient 10 below 25 and ten higer than 5 so there's no paintings here you basically word symatry and take equivalent quantities of each and every to realize the soundness so because of the fact the quantity you like is 1600ml you divide via 2 and take that from the two 800 in case you spot the symatry in this you keep alot of paintings

Answer 6

0.80x+0.40y=0.72(250)
80x+40y=18000
the other equation is x+y=250
multiplying by -40
-40y-40x=-10000
adding
40x=8000
x=200
so add 200 ml of 40% to 50 ml of 80%

Answer 7

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