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Luba_88 [7]
3 years ago
14

A science teacher has a supply of 5% hydrochloric acid and a supply of 65% hydrochloric acid(HC1). How much of each solution sho

uld 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.
Mathematics
1 answer:
Sphinxa [80]3 years ago
6 0
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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Answer:

The average rate of change on the given interval is 9/70

Step-by-step explanation:

Here, we are to find the average rate of change of the function on the given interval

We proceed as follows;

on an interval [a,b] , we can find the average rate of change using the formula;

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From the question;

a = 0

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Substituting the values, we have;

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= (42-15)/70/3

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= 27/70 * 1/3 = 9/70

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