Plate motion moves rock formations. Subduction moves rock down, below earths outer layer.
como se denomina el proceso utilizado para descomponer el agua
Answer:
Partial pressure of of CO₂ in the product mixture is 0,20atm
Explanation:
The balance equation is:
2CO(g) + O₂(g) → 2CO₂(g)
Total pressure of CO(g) and O₂(g) gases before reaction at 100,0°C and 1,0L is 1,50 atm. You can say:
X₁ + Y₁ = 1,50atm <em>(1)</em>
Where X₁ is initial partial pressure CO and Y₁ is initial partial pressure of O₂
After reaction partial pressures are:
X₂ = X₁ - 2n = 0; <em>2n = X₁</em>
Y₂ = Y₁ - n
Z₂ = 2n
Where Z₂ is final partial pressure of CO₂
After reaction pressure at 100,0°C and 1,0L is 1,40 atm, that means:
1,40 atm = (Y₂ + Z₂)
1,40 atm = Y₁ - n + 2n
1,40atm = Y₁ + n
1,40 atm = Y₁ + X₁/2 <em>(2)</em>
Replacing (1) in (2)
1,40 atm = 1,50atm - X₁ + X₁/2
-0,10 atm = - X₁/2
<em>0,20 atm = X₁</em>.
As 2n = X₁; 2n =<em> Z₂ = 0,20 atm</em>
I hope it helps!
Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.
On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where:
- P is the gas pressure.
- V is the volume that occupies.
- T is its temperature.
- R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
- n is the number of moles of the gas.
Then, in this case:
- P= 1 atm
- V= 44.1 L
- n= ?
- R= 0.082

- T= 0°C =273 K
Replacing in the expression for the ideal gas law:
1 atm× 44.1 L= n× 0.082
× 273 K
Solving:

n=1.97 moles
Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:
= 63.04 g ≈ <u><em>63 g</em></u>
Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
Learn more about the ideal gas law:
The lower the frequncy the longer the wavelength the longer the wavelength whoch would probably give you more energy.
if im incorrect im truly sorry