Answer : The volume of the bubble is, 625 mL
Explanation :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 2.4 atm
= final pressure of gas = 1.0 atm
= initial volume of gas = 250 mL
= final volume of gas = ?
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get:
Therefore, the volume of the bubble is, 625 mL
Answer:
The concentration of O2 will begin decreasing and The concentrations of CO2 and O2 will be equal.
Explanation:
Equilibrium occurs when the velocity of the formation of the products it's equal to the velocity of the formation of the reactants, thus the concentrations of the compounds remain constant.
Analyzing the information and the reaction given, we can notice that in equilibrium the rate (velocity) of formation of O2 (product) is equal to the rate of formation of CO2 (reactant).
As the CO2 and H2O are placed in the reaction, the Le Chateliêr's principle states that the equilibrium must shift to reestablish the equilibrium, thus, they must be consumed, and the concentration of O2 must increase.
As state above, in equilibrium, the concentrations didn't change, thus, the concentrations of CO2 and O2 will not change.
The concentrations of CO2 and O2 depends on the rate of the reaction and the initial quantities presented, so it's not possible to affirm they'll be equal.
Answer:
its a chemical formula, it has numbers and symbols
Answer:
A) E° = 4.40 V
B) ΔG° = -8.49 × 10⁵ J
Explanation:
Let's consider the following redox reaction.
2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)
We can write the corresponding half-reactions.
Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) E°red = 1.36 V
Anode (oxidation): 2 Li(s) → 2 Li⁺(aq) + 2 e⁻ E°red = -3.04
<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>
The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.
E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V
<em>B) Calculate the free energy ΔG° of the reaction.</em>
We can calculate Gibbs free energy (ΔG°) using the following expression.
ΔG° = -n.F.E°
where,
n are the moles of electrons transferred
F is Faraday's constant
ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J
