Answer:
The amount of energy required to break the ionic bonds in CaF2.
Aluminum has three oxidation states. The most common one is +3. The other two are +1 and +2. One +3 oxidation state for Aluminum can be found in the compound aluminum oxide, Al2O3.
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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The valence of helium is 0.
Answer:
0.188mol
Explanation:
Using the formula as follows:
mole = mass/molar mass
Molar mass of hypomanganous acid, H3MnO4 = 1(3) + 55 + 16(4)
= 3 + 55 + 64
= 122g/mol
Mass of H3MnO4 is given as 22.912 g
Hence;
moles = 22.912 ÷ 122
number of moles = 0.188mol