Answer:
<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>
The number of atoms of lead required is 1.73x10²³.
Explanation:
To find the number of atoms of lead we need to find first the volume of the plate:

<u>Where</u>:
A: is the surface area = 160
t: is the thickness = 0.002
<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>
Now, using the density we can find the mass:

Finally, with the Avogadros number (
) and with the atomic mass (A) we can find the number of atoms (N):
Hence, the number of atoms of lead required is 1.73x10²³.
I hope it helps you!
Answer:
(a) 0.243 m3/day
(b) 96 mg/l
(c) 0.426 m3/min
Explanation:
The sludge has an average solids concentration of 4 percent and considering TSS concentration in the influent of 240 mg/L then solids in sludge will be 0.04*240= 9.6 mg/L
Considering the average flow of 0.3 m3/s then mass of sludge per day will be given by 0.3*1000*9.6*60*60*24/1000000=248.832 kg/day
To get volume, considering specific gravity given as 1.025 and taking density of water as 1000 kg/m3 then density of sludge is 1025 kg/m3
Volume is mass/density hence 248.832/1025=0.2427629268292 m3/day
Approximately, the volume of sludge is 0.243 m3/day.
(b)
Efficiency of 60 percent is equivalent to 0.6
Efficiency=(influent concentration- flow rate)/influent concentration
0.6=(240-flow rate)/240
Flow rate= 96 mg/l
(c)
Cycle time= 0.243/0.57=0.4263157894736 m3/min
Rounded off, cycle time is 0.426 m3/day
Maybe To know short cuts on the long run. And it is a way of coding