Answer:
Kp = 0.049
Explanation:
The equilibrium in question is;
2 SO₂ (g) + O₂ (g) ⇄ 2 SO₃ (g)
Kp = p SO₃² / ( p SO₂² x p O₂ )
The initial pressures are given, so lets set up the ICE table for the equilibrium:
atm SO₂ O₂ SO₃
I 3.3 0.79 0
C -2x -x 2x
E 3.3 - 2x 0.79 - x 2x
We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of SO₂ and O₂ as follows:
p SO₂ = 3.3 -0.47 atm = 2.83 atm
p O₂ = 0.79 - (0.47/2) atm = .56 atm
Now we can calculate Kp:
Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )
Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.
The answer is b. radon-222. The alpha decay means that it will emit an alpha particle when decays. The alpha particle has two protons and two neutrons. So Radium(88) minus two protons will become Radon(86). And the atomic mass will become 226-4=222.
Answer:
Collisions between gas particles are elastic; there is no net gain or loss of kinetic energy.
Explanation:
When a gas is paced in a container, the molecules of the gas have little or no intermolecular interaction between them. There is a lot of space between the molecules of the gas.
The gas molecules move at very high speed and collide with each other and with the walls of container.
The collision of these particles with each other is perfectly elastic hence the kinetic energy of the colliding gas particles do not change.
Answer:- HBr is limiting reactant.
Solution:- The given balanced equation is:
From this equation, There is 2:6 mol or 1:3 mol ratio between Al and HBr. Since we have 8 moles of each, HBr is the limiting reactant as we need 3 moles of HBr for each mol of Al.
The calculations could be shown as:
= 24 mol HBr
From calculations, 24 moles of HBr are required to react completely with 8 moles of Al but only 8 moles of it are available. It clearly indicates, HBr is limiting reactant.
Answer:
2.4 mole of oxygen will react with 2.4 moles of hydrogen
Explanation:
As we know
1 liter = 1000 grams
2H2 + O2 --> 2H2O
Weight of H2 molecule = 2.016 g/mol
Weight of water = 18.01 gram /l
2 mole of oxygen react with 2 mole of H2
2.4 mole of oxygen will react with 2.4 moles of hydrogen
