Answer:
The answer to your question is Volume = 11.4 L
Explanation:
Data
Volume 1 = V1 = 6 L
Pressure 1 = P1 = 1 atm
Temperature 1 = T1 = 22°C
Volume 2 = V2 = ?
Pressure 2 = 0.45 atm
Temperature 2 = -21°C
Process
1.- Convert temperature (°C) to °K
T1 = 273 + 22 = 295°K
T2 = 273 + (-21) = 252°K
2.- Use the combined gas law to solve this problem
P1V1 / T1 = P2V2 / T2
-Solve for V2
V2 = P1V1T2 / T1P2
-Substitution
V2 = (6)(1)(252) / (295)(0.45)
- Simplification
V2 = 1512 / 132.75
- Result
V2 = 11.38 L
<h3>Answer:</h3>
89.6 L of O₂
<h3>Solution:</h3>
The balanced chemical equation is as,
CH₄ + 2 O₂ → CO₂ + 2 H₂O
As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,
44 g ( 1 mol) CO₂ is produced by = 44.8 L (2 mol) of O₂
So,
88 g CO₂ will be produced by = X L of O₂
Solving for X,
X = (88 g × 44.8 L) ÷ 44 g
X = 89.6 L of O₂
A circuit breaker must be replaced after too much current flowing through it causing it to melt.
Attraction between polar molecules
Option (a) is correct.
A reducing agent is the one which loses electrons to other substance and an oxidizing agent is one which accepts electrons.
Here, In

, Cr has oxidation number 6+ in the L.H.S of the equation, but on R.H.S its oxidation number is 0 i.e. it Cr has gained electrons such that total charge is 0.
And the oxidation state of Al in the left-hand side of equation is 0 and in right-hand side, it is +6.i.e. it has donated its electrons to Cr.
Hence, Cr is the oxidizing agent and Al is the reducing agent.