Question

a science teacher has a supply of 5% hydrochloric acid and a supply of 65% hydrochloric acid (HCI). how much of each solution should the teacher mix together to get 42 mL of 45% HCI for an experiment?

A- the teacher will need 14 mL of the 5% solution and 28 mL of the 65% solution.
B- the teacher will need 28 mL of the 5% solution and 14 mL of the 65% solution.
C- the teacher will need 14 mL of the 5% solution and 14 mL of the 65% solution.
D- the teacher will need 28 mL of the 5% solution and 28 mL of the 65% solution.

Answer 1

what do you mean your geting an error

Answer 2

@Euler271 the other equation which is
x+y=42
so solving x+y=42 and 0.05x + 0.65y = 0.45*42 will give u the reqd unique answer

Answer 3

check if it is
A- the teacher will need 14 mL of the 5% solution and 28 mL of the 65% solution.

Answer 4

matricked is right

Answer 5

sorry if it took me along time to reply

Answer 6

i'm gooing to be gone for tell 10:00 or later so sorry about that and i'll get back to you guys and girls

Answer 7

okay i'm back

Answer 8

teacher has 2 solutions.
1 = 5% hydrochloric acid.
1 = 65% hydrochloric acid.
how much of each solution to get 42 mL of 45% hydrochloric acid?

let x = number of mL of first solution.
let y = number of mL of second solution.

amount of HCI in first solution is .05*x
amount of HCI in second solution is .65*y

teacher wants 42 mL of 45% HCI.

amount of HCI in the solution that the teacher wants is .45 * 42 mL = 18.9 mL.

first equation is .05*x + .65*y = 18.9 mL of HCI.

first equation tells teacher how much HCI in each solution is required to total the amount of HCI required in the final solution.

second equation is x + y = 42 mL of total solution.

second equation tells teach how much total of each solution is required to get the proper mix.

these 2 equations need to be solved simultaneously to get the right answer.

the 2 equations are:

.05*x + .65*y = 18.9
x + y = 42

solve for y in the second equation to get y = 42 - x.

substitute for y in the first equation to get:

.05*x + .65*(42-x) = 18.9

first equation now has 1 unknown and can be solved.

solve for x.

simplify by removing parentheses in second equation to get:



.05*x + .65*42 - .65*x = 18.9

simplify further to get:

.05*x + 27.3 - .65*X = 18.9

subtract 27.3 from both sides of the equation to get:

.05*x - .65*x = - 8.4

combine like terms to get:

-.6*x = -8.4

divide both sides of the equation to get:

x = -8.4/-.6 = 14.

since x + y = 42, then 14 + y = 42.

subtract 14 from both sides of this equation to get:

y = 42 - 14 = 28

teacher has x = 14 and y = 28 which means that:

teacher needs 14 mL of the 5% solution and 28 mL of the 65% solution.

if teacher does that, then teacher gets:

.05*14 + .65*28 = .7 + 18.2 = 18.9 mL of HCI.

Answer looks good.

Answer is:

teacher needs 14 mL of 5% solution and 28 mL of 65% solution to get 42mL of 45% solution.

14 + 28 = 42 mL total solution.

.05*14 + .65*28 = 18.9 mL of HCI.

18.9 / 42 = .45 = 45% of HCI in final solution.

Answer 9

credit: http://www.algebra.com/algebra/homework/Systems-of-equations/Systems-of-equations.faq.question.349748.html

Answer 10

So the answer is A. :)
Tell me if this helped, and click "Best response" if it did :)

Answer 11

thank you this helped me alot

Answer 12

Click "best response" near my answer then :)
Thanks.

Answer 13

there you go man

Answer 14

well see you in the morning

Answer 15

Thanks :)

Answer 16

your welcome buddy

Answer 17

you now i haven't had this much help since a long time

Answer 18

bye-bye