1) <u>Stereo-selective (or enantioselective)</u> reactions form predominately or exclusively one enantiomer.
2) Epoxidation is the addition of a single oxygen atom to an alkene to form an epoxide.
3) <u>Hydrogenation (or reduction)</u> of an alkene forms an alkane by addition of H₂.
4) <u>Dihydroxylation</u> is the addition of two hydroxy groups to a double forming, a 1,2-diol or glycol.
5) <u>oxidative</u> cleavage of an alkene breaks both the σ and π bonds of the double bond to form two carbonyl groups.
6) <u>Regioselective</u> reactions form predominately or exclusively one constitutional isomer.
7) <u>Syn</u> dihydroxylation results when an alkene is treated KMnO4 or OsO4, where each reagent adds two oxygen atoms to the same side of the double bond.
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
It is beacuse the ions in the melted or aqueous ionic compound is mobile and can freely move through the fluid and conduct electricity.
The balanced chemical reaction is :
Number of moles of Na, 
.
Now, from balance chemical reaction we can see that 1 mole of oxygen reacts with 4 moles of sodium.
So, number of moles of oxygen are :
So, amount of oxygen required is :
Therefore, 5.08 gram of oxygen will react with 14.6 gram of sodium.
