Answer:
Part C: P2 = 0.30 atm
Part D: V1 = 16.22 L.
Explanation:
Part C:
Initial pressure (P1) = 2.67 atm
Initial volume (V1) = 5.54 mL
Final pressure (P2) =.?
Final volume (V2) = 49 mL
The final pressure (P2) can be obtained as follow:
P1V1 = P2V2
2.67 x 5.54 = P2 x 49
Divide both side by 49
P2 = (2.67 x 5.54)/49
P2 = 0.30 atm
Therefore, the final pressure (P2) is 0.30 atm
Part D:
Initial pressure (P1) = 348 Torr
Initial volume (V1) =?
Final pressure (P2) = 684 Torr
Final volume (V2) = 8.25 L
The initial volume (V1) can be obtained as follow:
P1V1 = P2V2
348 x V1 = 684 x 8.25
Divide both side by 348
V1 = (684 x 8.25)/348
V1 = 16.22 L
Therefore, the initial volume (V1) is 16.22 L
The bond angle in the Water Molecule is approx. 104.5 degrees. The methane molecule is approx. 109.5 degrees. The differences, is due to the force from the surrounding molecules and atoms. For example the Lone pairs of electrons.
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:
A) measuring mass of metal used in a reaction
Explanation:
Quantitive observations is data involving statistics and numerical values.
