Answer: The lead-containing reactant(s) consumed during recharging of a lead-acid battery is 
Explanation:
In lead acid battery, the anode is made up of lead and undergoes oxidation during discharging and cathode is made up of lead oxide and acts as cathode during discharging. The electrolyte used is dilute
.
Charging:
Cathode : reduction : 
Anode: oxidation : 
Overall reaction : 
The lead-containing reactant(s) consumed during recharging of a lead-acid battery is 
Answer: 45.983 g CBr₄
Explanation:
To convert from moles to grams, you know that we will need molar mass and Avogadro's number.
Avogadro's number: 6.022×10²³ molecules/mol
Molar mass: 331.627 grams/mol
Now that we have what we need, you can use these to solve for grams. 
Our final answer is 45.983 g CBr₄.
Answer:
K = 2.7x10⁻⁵ at 25ºC
Explanation:
A way to write Arrhenius equation is:
ln K = - Ea/R × (1/T) + lnA
If you graph ln K as Y and 1/T as X (Absolute temperature in K), the equation you will obtain is:
Y = -13815X +35.817
R² = 0.9927
(Taking the last k point as 0.0386) (ln 0.0386), <em>0.1386 has no sense</em>)
Your slope is -13815
-13815K = - Ea/R
-13815K×8.314J/molK = 114858J/mol = Ea
And your intercept =
lnA = 35.817
A = 3.59x10¹⁵
Now, you want to know rate constant at 25ºC = 298.15K. Replacing in the equation (Where Y is ln (activation energy) and X is 1/T):
Y = -13815X +35.817
Y = -13815(1/298.15K) +35.817
Y = -10.5187
lnK = -10.5187
<h3>K = 2.7x10⁻⁵ at 25ºC</h3>
Explanation:
Below are attachments containing the graph and solution.