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
C: 0.012 mol.
<h3>Explanation</h3>Start with 0.0020 moles of iodate ions .
How many moles of iodine will be produced?
converts to
in the first reaction. The coefficient in front of
is three times the coefficient in front of
. In other words, each mole of
will produce three moles of
. 0.0020 moles of
will convert to 0.0060 moles of
.
How many moles of thiosulfate ions are required?
reacts with
in the second reaction. The coefficient in front of
is twice the coefficient in front of
. How many moles of
does each mole of
consume? Two. 0.0060 moles of
will be produced. As a result,
moles of
will be needed.
Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.
