Equation: 2H₂ + O₂ → 2H₂O
Now, Given mass of Oxygen = 192 g
Molar mass of Oxygen = 16 g/mol
No. of moles in Oxygen = 16/192 = 0.0833
Now, for every mole of Oxygen, 2 mole of Hydrogen will form,
so, Number of moles of Hydrogen = 0.0833 * 2 = 0.167
Given mass = Number of Moles * Molar mass
Given mass = 0.167 * 2
m = 0.33 g
In short, Your Answer would be: 0.33 g
Hope this helps!
It depends on the process.
Like for example if the process is isothermal(temperature is constant), you can use,
PV = constant or P1V1 = P2V2 where P1V1 are initial conditions and P2V2 are final.
For adiabatic process,
PV^gamma = constant or P1V1 ^gamma = P2V2 ^gamma.
where gamma = Cp
------
Cv
Cp = specific heat at constant pressure and Cv = specific at constant volume.
Value of Gamma will be given in question.
Hope this helps!
Answer:
productivity and water depth
Explanation:
The productivity and the depth of water are both equally important as it directly affects the accumulation of biogenic sediments such as the siliceous ooze and calcareous ooze. In the equator and the coastal upwelling areas, and at the site of divergence of oceans, there occurs a high rate and amount of productivity, and these are considered to be the primary productivity.
The siliceous oozes are a good indicator of extensively high productivity in comparison to the carbonate oozes. The main reason behind this is that the silica can be easily dissolved in the surface water. On the other hand, the carbonates dissolve at a relatively lower ocean water depth, so there requires a high amount of surface productivity in order to allow these siliceous oozes to reach the ocean bottom.
Thus, the water depth and productivity, both are considered as the limiting factor in determining the accumulation of biogenic oozes.
Answer:
Explanation:
Hello, in this case, one could consider the undergoing chemical reaction as:
Thus, since 1.0 g of strontium carbonate is placed, the equilibrium equation takes the following form, excluding the solid-stated species and considering just the carbon dioxide as it is gaseous:
Hence, since at the beginning there is no carbon dioxide, its pressure at equilibrium equals Kp:
Which was clearly defined above.
Best regards.
