It would actually remain the same nothing will change.
Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
Answer:
This is due to the physical properties of the sample, since it affects the volume dispensed.
Explanation:
For example, in the case of very dense samples, selected samples to adhere to the surface of the tip, dispensing more slowly. In contrast, ethanol samples are less viscous and more volatile and are dispensed more rapidly. Some of the ways to minimize these inconveniences are the use of ultra low retention pipette tips, since they have a hydrophobic plastic additive that prevents the liquid from adhering to the inside of the tip.
Another way is to use the reverse pipetting.
This is not a scientific term. The scientific version is Nuclear ___. Depends on the context.
Answer:
a) 21.133 minutes for series
b) 84.54 minutes when split into 4
Explanation:
Data provided:
Total flow, V₀ = 45 MGD
total volume of each basin, V = 2500 m³
Now,
1 MGD = 3785.4118 m³/day
also,
1 day = 1440 minutes
thus,
45 MGD = 45 × 3785.4118 m³/day
or
45 MGD = 170343.5305 m³/day
and,
170343.5305 m³/day in 'm³/min'
= 170343.5305 / 1440
= 118.2941 m³/min.
Therefore,
The retention time, t =
or
The retention time, t =
or
The retention time, t = 21.13 min
Hence,
for series of tanks the retention time is 21.133 min.
Now,
on splitting the tanks in 4
the volumetric flow rate will be
=
= 29.57 m³/min.
Therefore,
The retention time =
or
The retention time, t =
or
The retention time, t = 84.54 min