Answer:
0.098 moles
Explanation:
Let y represent the number of moles present
1 mole of Ba(OH)₂ contains 2 moles of OH- ions.
Hence, 0.049 moles of Ba(OH)2 contains y moles of OH- ions.
To get the y moles, we then do cross multiplication
1 mole * y mole = 2 moles * 0.049 mole
y mole = 2 * 0.049 / 1
y mole = 0.098 moles of OH- ions.
1 mole of OH- can neutralize 1 mole of H+
Therefore, 0.098 moles of HNO₃ are present.
The reducing agent can approach the carbonyl face of camphor by forming a one carbon bridge (known as an exo attack) or a two carbon bridge (termed endo).
The two resultant stereoisomers are known as isoborneol and borneol (from exo attack) (from endo attack). Gas chromatography (GC) analysis may be used to calculate the ratio of each isomeric alcohol in the mixture. Unfortunately, IR analysis does not permit this.
The stereochemistry of the reaction is regulated in stiff cyclic compounds like camphor and norcamphor by protecting one side of the carbonyl group from the reagent's assault. The hydrogen atom is added to the endo side, creating the exo alcohol isoborneol, while the methyl groups on the one-carbon bridge of camphor screen the approach of the hydride from the "top" or exo side of the two-carbon bridge. You will be asked to guess the main isomeric alcohol created by the norcamphor hydride reduction later in the lab report.
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