Explanation:
(a) The given data is as follows.
Pressure on top (
) = 140 bar =
(as 1 bar =
)
Temperature =
= (15 + 273) K = 288 K
Density of gas = 


= 0.4548

=
= 
Hence, pressure at the natural gas-oil interface is
.
(b) At the bottom of the tank,

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]
= 
= 309.8 bar
Hence, at the bottom of the well at
pressure is 309.8 bar.
Increased upwelling in a coastal area results to more aquatic life. Upwelling is the process in which deep, cold water rises towards the surface. It is an oceanographic phenomenon that involves wind driven motion of the dense, cooler and usually nutrient-rich water towards the ocean surface replacing the warmer usually nutrient depleted surface water.
7. 1,2,2,2
8. 4,3,1
9. 1,4,1,2
10. 3,4,1,4
Answer:
The ideal gas law is expressed mathematically by the ideal gas equation as follows;
P·V = n·R·T
Where;
P = The gas pressure
V = The volume of the gas
n = The number of moles of the gas present
R = The universal gas constant
T = The temperature of the gas
A situation where the ideal gas law is exhibited is in the atmosphere just before rainfall
The atmospheric temperature of the area expecting rainfall drops, (when there is appreciable blockage of the Sun's rays by cloud covering) followed by increased wind towards the area, which indicates that the area was in a state of a low pressure, 'P', and or volume, 'V', or a combination of both low pressure and volume P·V
When the entry flow of air into the area is observed to have reduced, the temperature of the air in the area is simultaneously sensed to have risen slightly, therefore, the combination of P·V is seen to be proportional to the temperature, 'T', and the number of moles of air particles, 'n' in the area
Explanation:
<u>Given:</u>
Mass of H2O2 solution = 5.02 g
Mass of H2O2 = 0.153 g
<u>To determine: </u>
The % H2O2 in solution
<u>Explanation:</u>
Chemical reaction-
2H2O2(l) → 2H2O(l) + O2(g)
Mass % of a substance in a solution = (Mass of the substance/Mass of solution) * 100
In this case
% H2O2 = (Mass H2O2/Mass of solution)* 100 = (0.153/5.02)*100 = 3.05%
Ans: % H2O2 in the solution = 3.05%