Question

How much salt is in a mixture containing 10 gallons of an 11% salt solution and 6 gallons of a 7% salt solution?

Answer 1

- 10 gallons with 11% salt:
there is 0.11 to 1.0 salt, so there are 1.1 gallons salt
- 6 gallons with 7% salt:
there is 0.07 to 1.0 salt, and 1.0 ~ 6 gallons:
6 x 0.07 or 0.42 gallons of salt

- in total: 1.1 gallons + 0.42 gallons = 1.52 gallons of salt.

Answer 2

Thank you so much.
WHat would the percentage of pure water contained in this same mixture be?

Answer 3

first find the % of the salt:
we have 1.52 gallons salt in 10+6 gallons of water or
1.52 gall in 16 gall

to get a percentage we must compare with 100, not 16

Answer 4

then just subtract 1.52 from 16 gal?

Answer 5

we must make the denominator 100 while conserving the ratio

possible way achieving this:
1,52 * 100/16
-------------
16 * 100/16

Answer 6

want 100 in the denominator, so I multiply with 100
however there was already a 16?! so I also divide 16 to get rid of it
then only 100 is left in the denominator, like should be for %

we have to apply the same operations to the numerator: if we do this,
the ratio did not change, just the sizes

Answer 7

Karen is working in the chemistry lab and needs a solution that is 8% hydrochloric acid. She has two solutions that she can mix together to make the solution of the required strength. The first solution is 2% hydrochloric acid, and the other is 11% hydrochloric acid. How many kilograms of each solution should she mix to produce 3 kilograms of the solution of the required strength?

Answer 8

how would I do this

Answer 9

do you know the percentage of clear water for the prev one?

Answer 10

1,52 * 100/16
-------------
16 * 100/16

that's percentage of salt in new solution. percentage of pure water
is just 100% - above %

because they ask for PERCENTAGE and not parts of 16 so you can't
just subtract 1.52 from 16 g because that's not a real %

Answer 11

I see

Answer 12

I got 9,5% salt in the solution so it's 90,5% pure water in the solution

Answer 13

Karen is working in the chemistry lab and needs a solution that is 8% hydrochloric acid. She has two solutions that she can mix together to make the solution of the required strength. The first solution is 2% hydrochloric acid, and the other is 11% hydrochloric acid. How many kilograms of each solution should she mix to produce 3 kilograms of the solution of the required strength?

Answer 14

you could make an equation for that. is that a good way or are you supposed to solve this question with a different technique?

Answer 15

equation

Answer 16

OK

Answer 17

I really appreciate your help.

Answer 18

11 = x*(0.02) + y*(0.11) and also
x + y = 3

the reason to have 2 equations is: there could be more than one way
to get 8% hydrochloric acid, for example she could put double 11%
and also double the 2% solution, it would give the same concentration.

Answer 19

and that also makes things difficult....

Answer 20

no doubt

Answer 21

anyway it's possible to solve this SYSTEM OF EQUATIONS

Answer 22

but since it is all hypochloric acid, would I need x & y

Answer 23

x & y are the solutions of 11% and 2% basically like two plastic
containers, one of them has 11% and the other 2%

even though they all contain the same stuff, it's important to follow the
rule that you e.g. take 2 bottles of the one and 1 bottle of the other

Answer 24

otherwise it is hydrochloric solution however the concentration is wrong

Answer 25

SO how then would you solve this problem?

Answer 26

you can solve a system of equations by
1) subsitution
2) elimination

Answer 27

which one do you prefer? I don't know which one is better in this case
they both work

Answer 28

would I set it up like this?
.02x + .11(3 - x) = .24

Answer 29

yes, this is solving by substitution and that's a correct setup

Answer 30

where did you get .24 ?

Answer 31

from the 8% new solution, I took 3 kilograms x .08 = .24

Answer 32

so would the correct answer be 1 kilogram of the 8% and 2 kilograms of the 11%

Answer 33

we're already considering the 3kg in the y=3-x I believe

Answer 34

we can check if that's correct:

concentration
(1*8 + 2*11 )/3 = (22+8)/3 = 10% hydrochl.

Answer 35

this equation was wrong,
11 = x*(0.02) + y*(0.11) and also

it should be
0.08 = (0.02x + 0.11y)/(x+y)

Answer 36

so in the numerator we count the amount of salt that flows into the mix

and in the denominator we keep track of the total kg to be able to compute the %

Answer 37

and the % should be 8% or 8/100 = 0.08

Answer 38

right

Answer 39

ok so we have these two equations that must both be satisfied:
1.) 0.08 = (0.02x + 0.11y)/(x+y)
2.) x + y = 3

Answer 40

we can substitute y = 3-x in the first equation:

\[0.08 = \frac{(0.02x + 0.11[3-x])}{ (x+[3-x])}\]

Answer 41

and expand the substitute:
\[0.08 = \frac{(0.02x + 0.33-0.11x)}{ (x+3-x)}\]
we can combine like terms:
\[0.08 = \frac{(-0.09x + 0.33)}{ (3)}\]

Answer 42

and now, algebra - multiply both sides by 3:
\[0.24 = -0.09x + 0.33\]
and then move x to the left, move 0.24 to the right:
\[0.09x =+ 0.33 -0.24\]
combine the constants
\[0.09x =+ 0.09\]

Answer 43

so x=1

Answer 44

yes

and the fact we got such a whole number
indicates it might be correct

Answer 45

x+y = 3
y = 3-x

y= 3-1 =2

Answer 46

you already got that answer above
maybe while checking it

I confused 11% and 2% ??

Answer 47

x=2%, and y=11%

so if x=1 and y=2 then the concentration is:
(2% + 2*11%)/3 = (2+22 )/3 = 24/3 = 8%

Answer 48

ok you already had the right solution before all of the fractions
I just didn't check it correctly

at least now we have peer - reviewed the answer!!

Answer 49

she must put 1kg of the 2% solution
2 kg of the 11% solution

to get 3 kg of 8% solution

Answer 50

One more please

Answer 51

Susie is making a mixture of acetic acid and water, which forms vinegar. If she has a mixture of 100 ml of 2% acetic acid, how many ml of pure acetic acid should she add to produce the required 8% acetic acid vinegar? (Round to the nearest tenth.)

Answer 52

hello??

Answer 53

100 * 0.2 + x * 1.0 = 0.8/(100+x)

Answer 54

great

Answer 55

so x = .16 ??

Answer 56

then solve for x
100 * 0.2 + x * 1.0 = 0.8/(100+x)
(20 + x) *(100+x) = 0.8
2000 + 20x + 100 x + x^2 = 0.8
x^2 +120x + 2000 = 0.8
x^2 + 120x + 1999.2 = 0

Answer 57

you lost me there... so the answer is 0 ??

Answer 58

the eq was wrong

(100 * 0.2 + x * 1.0) /(100+x) = 0.8

Answer 59

(100 * 0.2 + x * 1.0) = 0.8 * (100+x)
20+x = 80+0.8x

Answer 60

20+0.2x = 80
0.2x = 60

2 x = 600
x = 300

Answer 61

thats clearly wrong

Susie is making a mixture of acetic acid and water, which forms vinegar. If she has a mixture of 100 ml of 2% acetic acid, how many ml of pure acetic acid should she add to produce the required 8% acetic acid vinegar? (Round to the nearest tenth.)

Answer 62

do you come up with .4

Answer 63

I see why, I used 0.2 for the 2% acid
I TREATED IT AS 20% acid

Answer 64

here with adjustment:
(100 * 0.02 + x * 1.0) /(100+x) = 0.08

Answer 65

that is what i was thinking
so did you come up with .16

Answer 66

you got a good idea about this stuff
I didn't come up with anything yet:

(100 * 0.02 + x * 1.0) = 0.08 * (100+x)
2 + x = 8+0.08x

Answer 67

haha

Answer 68

2 + x = 8+0.08x |subtract 0.08x from both sides
2 + 0.92 x = 8
0.92 x = 6 |(x100)
92 x = 600
x = 600/92

Answer 69

beleive it or not i am really trying to work it out, just working with the word problems throws me off

Answer 70

so I get 6,52 ml of pure acid are sufficient
to elevate 100ml of 2% upto 8%

Answer 71

gotcha

Answer 72

I guess thats plausable because we wanna add like 10% of acidity
so we add like a tenth of pure acid....

Answer 73

this will make an old man out of me

Answer 74

so we are not looking at 6.5

Answer 75

we are because the solution already had 2% and the real values are off by some small percentage due to the pure acid itself increasing volume

Answer 76

like if we had 100ml of 0% and we add 10ml of 100%
then we have 110 ml and it's slightly less than 10%

Answer 77

God bless you!! THanks for the help

Answer 78

thats why we need 6,5 ml and not just 6 ml
to get 2% upto 8%