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vivado [14]
3 years ago
9

Humans are most abundant during which geological era?

Chemistry
1 answer:
emmainna [20.7K]3 years ago
8 0

Answer:

Cenoizic

Precamborian

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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
hjlf

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
4 years ago
Which electron configuration represents a violation of the pauli exclusion principle?
Katarina [22]
<span>The electron configuration that represents a violation of the pauli exclusion principle is:

</span><span>1s: ↑↓ 
2s: ↑↑ 
2p: ↑</span>
The Pauli exclusion principle  refers to the quantum mechanical rule which expresses that at least two indistinguishable fermions (the particles with half-integer spin) can't involve a similar quantum state inside a quantum framework all the while.
7 0
3 years ago
Read 2 more answers
Plant cells have a large central vacuole, which animal cells lack. What function does this organelle perform?
Lapatulllka [165]
The central vacuole stores materials, wastes, and helps give the plant structure and support.

Hope this helps!
5 0
4 years ago
Read 2 more answers
Question Ask “ Which Part Of The Ecological Pyramid Is Incorrect.”
shepuryov [24]
Primary producers i believe
4 0
3 years ago
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The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate b
____ [38]

Answer:

Approximately 100 °C.

Explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

\Delta S_{vap}=\frac{\Delta H_{vap}}{T}

We can solve for the temperature as follows:

T=\frac{\Delta H_{vap}}{\Delta S_{vap}}

Thus, with the proper units, we obtain:

T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C

Hence, answer is approximately 100 °C.

Best regards.

6 0
3 years ago
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