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Papessa [141]
2 years ago
5

Indicate the compartment (A or B) that will increase in volume for each of the following pairs of solutions separated by a semi-

permeable membrane: A B a. 20% (m/v) starch b. 10% (m/v) albumin c. 0.5% (m/v) sucrose 10% (m/v) starch 2% (m/v) albumin 5% (m/v) sucrose​
Chemistry
1 answer:
avanturin [10]2 years ago
4 0

Based on the concentration of the solutions, the increase in volume is as follows:

  • the volume of the 20% (m/v) starch increases
  • the volume of the 10% (m/v) albumin increases
  • the volume of the 5% (m/v) sucrose increases

<h3>What is osmosis?</h3>

Osmosis is the movement of water molecules from a region of low solute concentration to a region of high solute concentration through a semipermeable membrane.

The movement of water molecules in the solution pairs is as follows:

  • the volume of the 20% (m/v) starch increases because water moves in from the 10% (m/v) starch solution.
  • the volume of the 10% (m/v) albumin increases because water moves in from the 2% (m/v) albumin
  • the volume of the 5% (m/v) sucrose increases because water moves in from the 0.5% (m/v) sucrose.

Therefore, the increase in volume depends on the concentration ofvthe solutions.

Learn more about osmosis at: brainly.com/question/2625460

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Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int
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Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.  

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.  

Million Gallons per day 1 MGD = 3785411.8 litre/day = 3785411.8 L/d

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Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.  

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C_f = \frac{F1*C1 +F2*C2}{F1 +F2}

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C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L

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Finally, we have to calculate the pounds of substance per day (Mp).  

We have the total flow F3 = F1 + F2 and the final concentration C_f. It is required to calculate per day, let's take a time of t = 1 day.  

F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg

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b) A total of 137.6 pounds pass a given spot downstream per day.

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