Answer:
monoxide
dioxide
trioxide
tetroxide
pentoxide
hexoxide
heptoxide
octoxide
nonoxide
Explanation:
Just the way it is. If there is an -a or -o infront of the oxide(or another substance that starts with vowel), the a is often dropped.
Answer:
1
Explanation:
4 HBr + O2 → 2H 20 + 2Br 2
...............
Convert mols to grams by multiplying grams of tin by the number of mols.
There are 119 grams per mol
119 x 11.8 = 1404 grams
Answer: Option (d) is the correct answer.
Explanation:
It is given that molecular formula is 
. Now, we will calculate the degree of unsaturation as follows.
Degree of unsaturation = 
= 
= 9 - 8 + 1
= 2
As the degree of unsaturation comes out to be 2. It means that this compound will contain one ring and one double bond.
Yes, this compound could be an alkyne as for alkyne D.B.E = 2.
But this compound cannot be a cycloalkane because for a cycloalkane D.B.E = 1 which is due to the ring only.
Thus, we can conclude that it is a cycloalkane is not a structural possibility for this hydrocarbon.
Option (i) would have the highest 2nd Ionization Energy.
Option (i) is Sodium.
Can be Written as 2, 8 , 1
For its 1st Ionization energy... It'd be extremely easy to remove that Electron cos its on the outermost shell.
Now After Removing that Electron...
Sodium's Electronic Configuration Reduces to that of Neon Which is 2, 8.
Neon has a very stable Octet.
It would take an ENORMOUS amount of energy to break its Octet stability... that is... Remove 1 electron from its Octet.
So
Option (i) [Sodium] has the highest 2nd Ionization Energy
