- Describe how you would make the salt from the reactants.
CuO + 2 HCl → CuCl₂ + H₂O
- Describe how you would purify the salt from the reaction mixture.
Filter the solution and let the product crystallize.
Explanation:
The reaction between copper oxide (CuO) and hydrochloric acid (HCl) will produce copper chloride (CuCl₂) and water:
CuO + 2 HCl → CuCl₂ + H₂O
-Describe how you would make the salt from the reactants
In a beaker which contain cooper oxide add the hydrochloric acid. To avoid working with concentrated hydrochloric acid, the acid may be diluted with water but make sure the add the stoechiometric amount (and a little bit of excess) in respect with the copper oxide.
-Describe how you would purify the salt from the reaction mixture.
Let the reaction proceed and then filter the solution to remove the unreacted cooper oxide.
Let the filtered solution, which contain copper chloride, water and unreacted hydrochloric acid, to stand undisturbed for several days. You may see the crystals growing from the solution. To speed up the process you may reduce the temperature.
Learn more about:
purifying compounds
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The given
ketone when reacted with base gave
enolate, the enolate formed due to loss of
methylene proton next to carbonyl group. Enolate when treated with
methyle Bromide gave
alpha substituted product.
Strong absorption around 1713 cm⁻¹ in IR spectrum confirms the presence of
Carbonyl group.
The product along with
¹H-NMR values is given below,
T = 20 % : 20 / 100 = 0.2
m1 = solute
m2 = Solvent
T = m1 / m1 + m2
0.2 = 500 g / 500 g + m2
0.2 * ( 500 + m2 ) = 500
0.2 * 500 + 0.2 m2 = 500
100 + 0.2 m2 = 500
0.2 m2 = 500 - 100
0.2 m2 = 400
m2 = 400 / 0.2
m2 = 2000 g of water
hope this helps!
To know this you pretty much do have to kind of memorize a few electronegativities. I don't recall ever getting a table of electronegativities on an exam.
From the structure, you have:
I remember the following electronegativities most because they are fairly patterned:
EN
H
=
2.1
EN
C
=
2.5
EN
N
=
3.0
EN
O
=
3.5
EN
F
=
4.0
EN
Cl
=
3.5
Notice how carbon through fluorine go in increments of
~
0.5
. I believe Pauling made it that way when he determined electronegativities in the '30s.
Δ
EN
C
−
Cl
=
1.0
Δ
EN
C
−
H
=
0.4
Δ
EN
C
−
C
=
0.0
Δ
EN
C
−
O
=
1.0
Δ
EN
O
−
H
=
1.4
So naturally, with the greatest electronegativity difference of
4.0
−
2.5
=
1.5
, the
C
−
F
bond is most polar, i.e. that bond's electron distribution is the most drawn towards the more electronegative compound as compared to the rest.
When the electron distribution is polarized and drawn towards a more electronegative atom, the less electronegative atom has to move inwards because its nucleus was previously favorably attracted to the electrons from the other atom.
That means generally, the greater the electronegativity difference between two atoms is, the shorter you can expect the bond to be, insofar as the electronegative atom is the same size as another comparable electronegative atom.
However, examining actual data, we would see that on average, in conditions without other bond polarizations occuring:
r
C
−
Cl
≈
177 pm
r
C
−
C
≈
154 pm
r
C
−
O
≈
143 pm
r
C
−
F
≈
135 pm
r
C
−
H
≈
109 pm
r
O
−
H
≈
96 pm
So it is not necessarily the least electronegativity difference that gives the longest bond.
Therefore, you cannot simply consider electronegativity. Examining the radii of the atoms, you should notice that chlorine is the biggest atom in the compound.
r
Cl
≈
79 pm
r
C
≈
70 pm
r
H
≈
53 pm
r
O
≈
60 pm
So assuming the answer is truly
C
−
C
, what would have to hold true is that:
The
C
−
F
bond polarization makes the carbon more electropositive (which is true).
The now more electropositive carbon wishes to attract bonding pairs from chlorine closer, thereby shortening the
C
−
Cl
bond, and potentially the
C
−
H
bond (which is probably true).
The shortening of the
C
−
Cl
bond is somehow enough to be shorter than the
C
−
C
bond (this is debatable).
