In the graphic, 195 represents the ______

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 $1 MGD = 3785411.8 litre/day = 3785411.8 L/d$

$F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d$

We have one flow of wastewater released into a stream.

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

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

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:

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

Replacing every value in L/d and mg/L

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

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

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.

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

After that, mg are converted to pounds.

$M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb$

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

4 0

3 years ago

When a rock falls from outer space all the way to the groung, its called a?

goblinko [34]

It's called a meteorite.

6 0

3 years ago

What is 20 times 200000000

iren [92.7K]

Answer: 4000000000

because 20X200000000=4000000000

4 0

2 years ago

A small hole in the wing of a space shuttle requires a 20.7-cm² patch.

cricket20 [7]

The Patch's area of the space shuttle in km² is 2.07 × 10⁻⁹ km²

Given, that a space shuttle requires a 20.7 cm² patch

We have to convert the patch's area from cm² into km².

Unit conversion is a method in which we multiply or divide with a particular numerical factor and then finally round off to the nearest significant digits.

Patch area of the space shuttle is 20.7 cm²

1 cm = 0.00001 km

or, 1 cm² = (0.00001 km)²

or, 1 cm² = 10⁻¹⁰km²

20.7 cm² = 20.7 × 10⁻¹⁰km²

20.7 cm² = 2.07 × 10⁻⁹ km²

The patch area in square kilometers is 2.07 × 10⁻⁹ km²

To learn more about unit conversion, visit: brainly.com/question/11543684

#SPJ4

8 0

1 year ago

What process removes carbon dioxide from the ocean?

Katen [24]

Answer:

B. decay of dead marine organisms

Explanation:

When the temperature is low, carbon dioxide is captured by the oceans, and when the temperature is high, it is released by the oceans into the atmosphere. At sea, carbon dioxide feeds phytoplankton.

Most of the carbon dioxide consumed by plant plankton (phytoplankton) returns to the atmosphere when this phytoplankton dies or is consumed, but a portion is deposited in the ocean floor sediments when these small particles sink. This process is called a "biological bomb" because carbon dioxide is transported from the atmosphere to the ocean floor.

8 0

3 years ago

In the graphic, 195 represents the _______.