Given:
P1 = 13.0 atm
T1 = 20 °C
T2 = 102 °C
Required:
P2 of oxygen
Solution:
At constant volume,
we can apply Gay-Lussac’s law of pressure and temperature relationship
P1/T1=P2/T2
(13.0 atm) / (20 °C)
= P2 / (102 °C)
P2 = 66.3 atm
The answer is not in the choices given.
Answer:
qualitative data : quantitative data :
circular in shape 75 colonies ...
stained purple 200 purple ..
spreads across plate 55 colonies ...
Explanation:
i got it right :) .
Answer:
b. 0,99atm
c. Answer is in the explanation
d. Answer is in the explanation
Explanation:
b. Using Gay-Lussac's law:
P₁T₂ = P₂T₁
P₁: 0,70 atm; T₂: 425K; P₂: ??; T₁: 299K
0,70atm×425K / 299K = <em>0,99 atm</em>
c. Using kinetic molecular theory, the increasing of temperature increases the kinetic energy of gas particles and if kinetic energy increases, the pressure increases. That means the increasing of temperature increases the pressure in the system.
d. Now, the increases in kinetic energy of gases increase the collisions betwen particles. As these intermolecular forces that are not taken into account in ideal gas law, the observed pressure will be different to the pressure predicted by ideal gas law.
I hope it helps!
<u>Given:</u>
Moles of Al = 0.4
Moles of O2 = 0.4
<u>To determine:</u>
Moles of Al2O3 produced
<u>Explanation:</u>
4Al + 3O2 → 2Al2O3
Based on the reaction stoichiometry:
4 moles of Al produces 2 moles of Al2O3
Therefore, 0.4 moles of Al will produce:
0.4 moles Al * 2 moles Al2O3/4 moles Al = 0.2 moles Al2O3
Similarly;
3 moles O2 produces 2 moles Al2O3
0.4 moles of O2 will yield: 0.4 *2/3 = 0.267 moles
Thus Al will be the limiting reactant.
Ans: Maximum moles of Al2O3 = 0.2 moles
Answer:
- Volume is a physical property of an object.
- One unit of volume is the milliliter.
- Liquids and solids have constant volumes.
Explanation:
