Answer:
The empirical formula is the simplest form;
Given:
Oxygen O at 94.1% and
H at 5.9%
Assume 100grams.
94% = 0.941 x 100gm. = 94.1 gm x 1mole/16gm. = 5.88 moles of O
5.9% = 0.059 x 100gm. = 5.9gm. X 1moleH/1.002gm. = 5.88 moles of H
There is one mole of O for each mole of H so the empirical formula is 
and written as OH.
You use a nucleus with inflection on the arm processor.
Answer:
CH3
|
CH3- C H -CH2-CH2- CH - CH2-CH2-CH3
|
CH
/ \
CH3 CH3
Explanation:
Octan
C-C-C-C-C-C-C-C
Metyl
CH3 -
Isopropyl
CH3
/
- CH
\
CH3
2-metil-5-isopropiloctan
CH3
|
CH3- C H -CH2-CH2- CH - CH2-CH2-CH3
|
CH
/ \
CH3 CH3
2 moles of sodium hydroxide will be needed.
<h3><u>Explanation</u>:</h3>
Sodium hydroxide is a compound which is a base and nitric acid is the acid. The formula of the nitric acid is HNO3 and that of sodium hydroxide is NaOH.
The reaction between them are
NaOH +HNO3 =NaNO3 +H2O.
So here we can see that 1 mole of sodium hydroxide reacts with 1 mole of nitric acid to produce 1 mole of sodium nitrate and 1 mole of water.
So for 2 moles of nitric acid, 2 moles of sodium hydroxide will be required.
