Answer:
Explanation:
From the given information:
The equation for the reaction can be represented as:
The I.C.E table can be represented as:
2SO₂ O₂ 2SO₃
Initial: 14 2.6 0
Change: -2x -x +2x
Equilibrium: 14 - 2x 2.6 - x 2x
However, Since the amount of sulfur trioxide gas to be 1.6 mol.
SO₃ = 2x,
then x = 1.6/2
x = 0.8 mol
For 2SO₂; we have 14 - 2x
= 14 - 2(0.8)
= 14 - 1.6
= 12.4 mol
For O₂; we have 2.6 - x
= 2.6 - 1.6
= 1.0 mol
Thus;
[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,
[O₂] = 1/50 = 0.02 M ,
[SO₃] = 1.6/50 = 0.032 M
Kc = [SO₃]² / [SO₂]² [O₂]
= ( 0.032²) / ( 0.248² x 0.02)
= 0.8325
Recall that; the equilibrium constant for the reaction = 0.8325;
If we want to find:
Then:
Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.
Answer:
The answer is 2.20 M
Explanation:
This is because ammonia has a pH of 11.8 and if you take 14-11.8 it equals 2.2 so the answer is 2.20 M
Half life is the time taken by a radioactive isotope to decay by half its original mass. In this case, the halflife of the radioactive isotope is 5000 years.
Initially the mass is 100 %; thus the mass that will be left will be given by;
New mass = Original mass × (1/2)^n where n is the number of half lives;
n = 10000/5000 = 2
New mass = 100% ×(1/2)^2
= 100 % × 1/4
= 25%
Therefore; the mass left after 10000 years is 25% or 1/4 of the original mass.
We found cesium, strontium, aluminum, sulfur, chlorine, and fluorine on the periodic table. Cesium is the farthest left and the lowest, while fluorine is the farthest right and the highest, so we know they have the highest metallic character and the lowest metallic character, respectively.
Yeah...it should be out of your system by then
