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Finger [1]
3 years ago
7

Clayton will mix xxx milliliters of a 10\%10%10, percent by mass saline solution with yyy milliliters of a 20\%20%20, percent by

mass saline solution in order to create an 18\%18%18, percent by mass saline solution. The equation above represents this situation. If Clayton uses 100100100 milliliters of the 20\%20%20, percent by mass saline solution, how many milliliters of the 10\%10%10, percent by mass saline solution must he use?
Chemistry
1 answer:
avanturin [10]3 years ago
3 0

Answer:

25 mL of the 10% solution.

Explanation:

<em>Clayton will mix x milliliters of a 10%, percent by mass saline solution with y milliliters of a 20%, percent by mass saline solution in order to create an 18%, percent by mass saline solution.</em> This can be described with the following equation:

0.1 x + 0.2 y = 0.18 (x + y)   [1]

<em>If Clayton uses 100 milliliters of the 20%, percent by mass saline solution, how many milliliters of the 10%, percent by mass saline solution must he use?</em>

If y = 100 mL, what should be the value of x?

Let's solve [1] for "x" and replace with the value of "y".

0.1 x + 0.2 y = 0.18 (x + y)

0.1 x + 0.2 y = 0.18 x + 0.18 y

0.02 y = 0.08 x

x = 0.25 y = 0.25 × 100 mL = 25 mL

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Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co
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Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as p_{d} and the proportion of air that is by-passed as p_{bp}, being p_{d}+p_{bp}=1.

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Replacing the first equation in the second one we have

0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

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100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

7 0
3 years ago
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