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Vlad [161]
Answer:
Explanation:
you need more information x
Answer:
overflow rate 20.53 m^3/d/m^2
Detention time 2.34 hr
weir loading 114.06 m^3/d/m
Explanation:
calculation for single clarifier
![sewag\ flow Q = \frac{12900}{2} = 6450 m^2/d](https://tex.z-dn.net/?f=sewag%5C%20%20flow%20Q%20%3D%20%5Cfrac%7B12900%7D%7B2%7D%20%3D%206450%20m%5E2%2Fd)
![surface\ area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2](https://tex.z-dn.net/?f=surface%5C%20%20area%20%3D%5Cfrac%7Bpi%7D%7B4%7D%5Ctimes%20diameter%20%5E2%20%3D%20%5Cfrac%7Bpi%7D%7B4%7D%5Ctimes%2020%5E2)
![surface area = 314.16 m^2](https://tex.z-dn.net/?f=surface%20area%20%3D%20314.16%20m%5E2)
volume of tank![V = A\times side\ water\ depth](https://tex.z-dn.net/?f=%20V%20%20%3D%20A%5Ctimes%20side%5C%20water%5C%20depth)
![=314.16\times 2 = 628.32m^3](https://tex.z-dn.net/?f=%3D314.16%5Ctimes%202%20%3D%20628.32m%5E3)
![Length\ of\ weir = \pi \times diameter of weir](https://tex.z-dn.net/?f=Length%5C%20of%5C%20%20weir%20%3D%20%5Cpi%20%5Ctimes%20diameter%20of%20weir)
![= \pi \times 18 = 56.549 m](https://tex.z-dn.net/?f=%20%3D%20%5Cpi%20%5Ctimes%2018%20%3D%2056.549%20m)
overflow rate =![v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2](https://tex.z-dn.net/?f=%20v_o%20%3D%20%5Cfrac%7Bflow%7D%7Bsurface%5C%20area%7D%20%3D%20%5Cfrac%7B6450%7D%7B314.16%7D%20%3D%2020.53%20m%5E3%2Fd%2Fm%5E2)
Detention time![t_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr](https://tex.z-dn.net/?f=%20t_d%20%3D%20%5Cfrac%7Bvolume%7D%7Bflow%7D%20%3D%20%5Cfrac%7B628.32%7D%7B6450%7D%20%5Ctimes%2024%20%3D%202.34%20hr)
weir loading![= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bflow%7D%7Bweir%5C%20length%7D%20%3D%20%5Cfrac%7B6450%7D%7B56.549%7D%20%3D%20114.06%20m%5E3%2Fd%2Fm)
Answer:
Check the explanation
Explanation:
A vending machine controller is that type of machine that comes with a single serial port on the same chip as the programmable processor. The controller comprises of a port arbitrator that selectively attaches or links one of a number of serially communicating devices to this single serial port.
Kindly check the attached image to get the step by step explanation to the above question.
Answer:
a. 0.4544 N
b. ![5.112 \times 10^{-5 M}](https://tex.z-dn.net/?f=5.112%20%5Ctimes%2010%5E%7B-5%20M%7D)
Explanation:
For computing the normality and molarity of the acid solution first we need to do the following calculations
The balanced reaction
![H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O](https://tex.z-dn.net/?f=H_2SO_4%20%2B%202NaOH%20%3D%20Na_2SO_4%20%2B%202H_2O)
![NaOH\ Mass = Normality \times equivalent\ weight \times\ volume](https://tex.z-dn.net/?f=NaOH%5C%20Mass%20%3D%20Normality%20%5Ctimes%20equivalent%5C%20weight%20%5Ctimes%5C%20volume)
![= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL](https://tex.z-dn.net/?f=%3D%200.3200%20%5Ctimes%2040%20g%20%5Ctimes%2021.30%20mL%20%5Ctimes%20%201L%2F1000mL)
= 0.27264 g
![NaOH\ mass = \frac{mass}{molecular\ weight}](https://tex.z-dn.net/?f=NaOH%5C%20mass%20%3D%20%5Cfrac%7Bmass%7D%7Bmolecular%5C%20weight%7D)
![= \frac{0.27264\ g}{40g/mol}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.27264%5C%20g%7D%7B40g%2Fmol%7D)
= 0.006816 mol
Now
Moles of
needed is
![= \frac{0.006816}{2}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.006816%7D%7B2%7D)
= 0.003408 mol
![Mass\ of\ H_2SO_4 = moles \times molecular\ weight](https://tex.z-dn.net/?f=Mass%5C%20of%5C%20H_2SO_4%20%3D%20moles%20%5Ctimes%20molecular%5C%20weight)
![= 0.003408\ mol \times 98g/mol](https://tex.z-dn.net/?f=%3D%200.003408%5C%20mol%20%5Ctimes%2098g%2Fmol)
= 0.333984 g
Now based on the above calculation
a. Normality of acid is
![= \frac{acid\ mass}{equivalent\ weight \times volume}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bacid%5C%20mass%7D%7Bequivalent%5C%20weight%20%5Ctimes%20volume%7D)
![= \frac{0.333984 g}{49 \times 0.015}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.333984%20g%7D%7B49%20%5Ctimes%200.015%7D)
= 0.4544 N
b. And, the acid solution molarity is
![= \frac{moles}{Volume}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bmoles%7D%7BVolume%7D)
![= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.003408%20mol%7D%7B15%5C%20mL%20%5Ctimes%20%201L%2F1000%5C%20mL%7D)
= 0.00005112
=![5.112 \times 10^{-5 M}](https://tex.z-dn.net/?f=5.112%20%5Ctimes%2010%5E%7B-5%20M%7D)
We simply applied the above formulas
Answer:
thanks thanks thanks thanks
Explanation: