Answer:
The volume of the balloon will be 5.11L
Explanation:
An excersise to solve with the Ideal Gases Law
First of all, let's convert the pressure in mmHg to atm
1 atm = 760 mmHg
760 mmHg ___ 1 atm
755.4 mmHg ____ (755.4 / 760) = 0.993 atm
922.3 mmHg ____ ( 922.3 / 760) = 1.214 atm
T° in K = 273 + °C
28.5 °C +273 = 301.5K
26.35°C + 273= 299.35K
P . V = n . R .T
First situation: 0.993atm . 6.25L = n . 0.082 . 301.5K
(0.993atm . 6.25L) / 0.082 . 301.5 = n
0.251 moles = n
Second situation:
1.214 atm . V = 0.251 moles . 0.082 . 301.5K
V = (0.251 moles . 0.082 . 301.5K) / 1.214 atm
V = 5.11L
Answer: Final temperature of the gas will be 330 K.
Explanation:
Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.
(At constant volume and number of moles)
where,
= initial pressure of gas = 1.00 atm
= final pressure of gas = 1.13 atm
= initial temperature of gas = 
K
= final temperature of gas = ?
Therefore, the final temperature of the gas will be 330 K.
gases become more soluble in liquids as the temperature gets higher
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Answer:
Thermal energy always moves from where there is<u> more </u>of it to where there is<u> less </u>of it until <u>distributed</u>.
Explanation:
Its true, just don't know if its correct for your assignment
Answer:
-122 J/K
Explanation:
Let's consider the following balanced reaction.
N₂(g) + 2 O₂(g) ⇒ 2 NO₂(g)
We can calculate the standard reaction entropy (ΔS°) using the following expression.
ΔS° = Σ ηp × Sf°p - Σ ηr × Sf°r
where,
- η: stoichiometric coefficients of products and reactants
- Sf°r: entropies of formation of products and reactants
ΔS° = 2 mol × 240.06 J/K.mol - 1 mol × 191.61 J/K.mol - 2 mol × 205.14 J/K.mol
ΔS° = -121.77 J/K ≈ -122 J/K
