Answer:
7.37 mL of KOH
Explanation:
So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,
HNO3 + KOH → KNO3 + H2O
Step 1 : The moles of HNO3 here can be calculated through the given molar mass ( 0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.
Mol of NHO3 = 0.140 M 
30 / 1000 L = 0.140 M 0.03 L = .0042 mol
Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,
0.0042 mol HNO2 ( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH
From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,
Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).
Volume of KOH = 0.0042 mol ( 1 L / 0.570 mol ) ( 1000 mL / 1 L ) = 7.37 mL of KOH
Answer:
(3S)-2-chloro-2,3-dimethylpentane is produced exclusively.
Explanation:
Electrophilic addition to (3S)-2,3-dimethylpent-1-ene proceeds through a carbocationic intermediate.
In the first step, 
adds onto double bond to produce more stable tertiary carbocation. (protonation)
In the second step, 
adds onto carbocation to produce (3S)-2-chloro-2,3-dimethylpentane exclusively.(nucleophilic addition)
So, option (d) is correct.
Answer:
The potential wrt. calomel is 1.254 V
Explanation:
Given:
Potential wrt. silver chloride 
V
Potential wrt. saturated silver chloride 
V
Potential wrt. SCE 
V
Now potential wrt. hydrogen is given by,
V
And we find for potential wrt. calomel,
potential wrt. hydrogen + potential wrt. SEC
V
Therefore, the potential wrt. calomel is 1.254 V
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