To be honest I do not really have a clue
Answer:
71.372 g or 0.7 moles
Explanation:
We are given;
- Moles of Aluminium is 1.40 mol
- Moles of Oxygen 1.35 mol
We are required to determine the theoretical yield of Aluminium oxide
The equation for the reaction between Aluminium and Oxygen is given by;
4Al(s) + 3O₂(g) → 2Al₂O₃(s)
From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.
Therefore;
1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen
1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium
Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.
4 moles of aluminium reacts to generate 2 moles aluminium oxide.
Therefore;
Mole ratio Al : Al₂O₃ is 4 : 2
Thus;
Moles of Al₂O₃ = Moles of Al × 0.5
= 1.4 moles × 0.5
= 0.7 moles
But; 1 mole of Al₂O₃ = 101.96 g/mol
Thus;
Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol
= 71.372 g
It’s D prokaryote - it best describes cell two
Hi! the english version of the given question is "A helium balloon is inflated to the volume of 0.045 m3 at a temperature of 2 ° C. If the balloon is cooled to -12 ° C. What will its new volume be? Consider that the pressure does not vary". The answer is given in english.
Answer:
The new volume of balloon is
.
Explanation:
Let's assume the helium gas inside the balloon behaves ideally.
The total number of moles of helium gas inside the balloon remains constant.
Applying combined gas law, we get:
![\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BP_%7B1%7DV_%7B1%7D%7D%7BT_%7B1%7D%7D%3D%5Cfrac%7BP_%7B2%7DV_%7B2%7D%7D%7BT_%7B2%7D%7D)
Where:
are the initial and final pressure of the balloon respectively.
are the initial and final volume of the balloon respectively.
are the initial and final temperature in the kelvin scale respectively.
Given:
![P_{1}=P_{2}](https://tex.z-dn.net/?f=P_%7B1%7D%3DP_%7B2%7D)
![V_{1}=0.045\;m^{3}](https://tex.z-dn.net/?f=V_%7B1%7D%3D0.045%5C%3Bm%5E%7B3%7D)
![T_{1}=(273+2)\;K=275\;K](https://tex.z-dn.net/?f=T_%7B1%7D%3D%28273%2B2%29%5C%3BK%3D275%5C%3BK)
![T_{2}=(273-12)\;K=261\;K](https://tex.z-dn.net/?f=T_%7B2%7D%3D%28273-12%29%5C%3BK%3D261%5C%3BK)
Substituting the above values, we get:
![\frac{(0.045\;m^{3})}{275\;K}=\frac{V_{2}}{261\;K}](https://tex.z-dn.net/?f=%5Cfrac%7B%280.045%5C%3Bm%5E%7B3%7D%29%7D%7B275%5C%3BK%7D%3D%5Cfrac%7BV_%7B2%7D%7D%7B261%5C%3BK%7D)
![\Rightarrow V_{2}=0.043\;m^{3}](https://tex.z-dn.net/?f=%5CRightarrow%20V_%7B2%7D%3D0.043%5C%3Bm%5E%7B3%7D)