<span>Fractionation - to expand from my answer to a previous question fractionation can be used to separate out more than one component based on a specific property. You still wanted to use Distillation on the other answer. Fractionation on this one.</span>
Answer:
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
Explanation:
When we balance a chemical equation, what we are trying to do is to achieve the same number of atoms for each element on both sides of the arrow. On the right of the arrow is where we can find the products, while the reactants are found on the left of the arrow.
We usually balance O and H atoms last.
AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 1
Cl --- 3
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
2 AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of As atoms is now balanced.
2 AsCl₃ + 3 H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of S atoms is now equal on both sides.
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
The equation is now balanced.
Answer : The volume for 6.0m HCl solution required = 62.5 ml
Solution : Given,
Initial concentration of HCl solution = 6.0m
Final concentration of HCl solution = 1.5m
Final volume of HCl solution = 250 ml
Initial volume of HCl solution = ?
Formula used for dilution is,
where,
= initial concentration
= final concentration
= initial volume
= final volume
Now put all the given values in above formula, we get the initial volume of HCl solution.
= 62.5 ml
Therefore, the volume for 6.0m HCl solution required = 62.5 ml
Answer:
CH₄ - 162 ⁸C
CH₃CH₃ -88.5 ⁸C
(CH₃)₂ CHCH₂CH₃ 28 ⁸C
CH₃3(CH2)₃CH₃ 36 ⁸C
CH₃OH 64.5 ⁸C
CH₃CH₂OH 78.3 ⁸C
CH₃CHOHCH₃ 82.5 ⁸C
C₅H₉OH 140 ⁸C
C₆H₅CH₂OH 205 ⁸C
HOCH₂CHOHCH₂OH 290 ⁸C
Explanation:
To answer this question we need first to understand that for organic compounds:
a. Non polar compounds have lower boiling points than polar ones of similar structure and molecular weight.
b. Boiling points increase with molecular weight. In alkane compounds if we compare isomers, the straight chain isomer will have a higher boiling point than the branched one (s) because of London dispersion intermolecular forces.
a. The introduction of hydroxyl groups increase the intermolecular forces and hence the boiling points because the electronegative oxygen, and, more importantly the presence of hydrogen bonds.
Considering the observations above, we can match the boiling points as follows:
CH₄ - 162 ⁸C
CH₃CH₃ -88.5 ⁸C
(CH₃)₂ CHCH₂CH₃ 28 ⁸C
CH₃3(CH2)₃CH₃ 36 ⁸C
CH₃OH 64.5 ⁸C
CH₃CH₂OH 78.3 ⁸C
CH₃CHOHCH₃ 82.5 ⁸C
C₅H₉OH 140 ⁸C
C₆H₅CH₂OH 205 ⁸C
HOCH₂CHOHCH₂OH 290 ⁸C
Note: There was a mistake in the symbols used for the 162 and 88.5 values which are negative and correspond to the common gases methane and ethane
Answer: C. Some are metals and some are nonmetals.
Explanation:
