Bonding and bond length in benzene and explain the pi and sigma bonds
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<u>Answer:</u> The mole ratio of H : O in ammonium nitrate is 4 : 3.
<u>Explanation:</u>
We are given a compound named ammonium nitrate having formula
There are 3 elements in this compound which are nitrogen, hydrogen and oxygen.
To calculate the mole ratio, we write the ratio of their subscripts. For this compound, it is:
The mole ratio of H and O for this compound is 4 : 3.
Answer:
2.387 mol/L
Explanation:
The reaction that takes place is:
- 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
First we <u>calculate how many moles of each reagent were added</u>:
- HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl
- Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂
460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.
Now we <u>calculate how many moles of Ba(OH)₂ reacted</u>, by c<em>onverting the total number of HCl moles to Ba(OH)₂ moles</em>:
- 203.85 mmol HCl *
= 101.925 mmol Ba(OH)₂
This means the remaining Ba(OH)₂ is:
- 460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂
There are two OH⁻ moles per Ba(OH)₂ mol:
- OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻
Finally we <u>divide the number of OH⁻ moles by the </u><u><em>total</em></u><u> volume</u> (100 mL + 200 mL):
- 716.15 mmol OH⁻ / 300.0 mL = 2.387 M
So the answer is 2.387 mol/L
Answer:
The percent composition of fluorine is 65.67%
Explanation:
Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.
That is, the percentage composition is the percentage by mass of each of the elements present in a compound.
The calculation of the percentage composition of an element is made by:
In this case, the percent composition of fluorine is:
percent composition of fluorine= 65.67%
<u><em>The percent composition of fluorine is 65.67%</em></u>