Ascience teacher has a supply of 5% hydrochloric acid and a supply of 65% hydrochloric acid(hc1). how much of each solution should the teacher mix together to get 42 ml of 45% hc1 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 5g solution and 14 ml of the 65% solution. c. the teacher will need 14 of ml of both solutions. d. the teacher will need 28 ml of both solutions.

Assume that the amount needed from the 5% solution is x and that the amount needed from the 65% solution is y.

We are given that, the final solution should be 42 ml, this means that:
x + y = 42 > equation I
This can also be written as:
x = 42-y > equation II

We are also given that the final concentration should be 45%, this means that:
5% x + 65% y = 45% (x+y)
0.05x + 0.65y = 0.45(x+y)

We have x+y = 42 from equation I, therefore:
0.05x + 0.65y = 0.45(42)
0.05x + 0.65y = 18.9 > equation III

Substitute with equation II in equation III as follows:
0.05x + 0.65y = 18.9
0.05(42-y) + 0.65y = 18.9
2.1 - 0.05y + 0.65y = 18.9
0.6y = 18.9 - 2.1
0.6y = 16.8
y = 28 ml

Substitute with y in equation II to get x as follows:
x = 42-y
x = 42 - 28
x = 14 ml

Based on the above calculations:
amount of 5% solution = x = 14 ml
amount of 65% solution = y = 28 ml
The correct choice is:
The teacher will need 14 mL of the 5% solution and 28 mL of the 65% solution.

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< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax   (given)ax ≅ ax (common side)

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)