Answer:
The answer to your question is 308.1 kPa
Explanation:
Data
Volume 1 = V1 = 4.75 l
Pressure 1 = P1 = 120 kPa
Volume 2 = V2 = 1.85 l
Pressure 2 = ?
Process
To solve this problem use Boyle's law.
P1V1 = P2V2
- Solve for P2
P2 = P1V1 / V2
- Substitution
P2 = (120 x 4.75) / 1.85
-Simplification
P2 = 570 / 1.85
-Result
P2 = 308.1 kPa
One of the most powerful laws in physics is the law of momentum conservation. ... For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.
Answer:
The entropy decreases.
Explanation:
The change in the standard entropy of a reaction (ΔS°rxn) is related to the change in the number of gaseous moles (Δngas), where
Δngas = n(gaseous products) - n(gaseous reactants)
- If Δngas > 0, the entropy increases
- If Δngas < 0, the entropy decreases.
- If Δngas = 0, there is little or no change in the entropy.
Let's consider the following reaction.
2 H₂(g) + O₂(g) ⟶ 2 H₂O(l)
Δngas = 0 - 3 = -3, so the entropy decreases.
There are 0.000882 gram in 882 ug