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).
The pH of the unknown solution is 3.07.
<u>Explanation:</u>
<u>1.Find the cell potential as a function of pH</u>
From the Nernst Equation:
Ecell=E∘cell−RT /zF × lnQ
where
R denotes the Universal Gas Constant
T denotes the temperature
z denotes the moles of electrons transferred per mole of hydrogen
F denotes the Faraday constant
Q denotes the reaction quotient
Substitute the values,
E∘cell=0 lnQ=2.303logQ
E0cell=−2.30/RT /zF × log Q
Solving the equation,
<u>2. Find the Q value</u>
Q=[H+]2prod pH₂, product/ [H+]2reactpH₂, reactant
Q=[H+]^2×1/1×1=[H+]2
Taking the log
logQ= log[H+]^2=2log[H+]=-2pH
From the formula,
Ecell=−2.303RT /zF× logQ
E cell= 2.303 × 8.314 CK mol (inverse) × 298.15
K × 2pH /2×96 485 C⋅mol
( inverse)
E cell= 0.0592 V × pH
<u>3. Finding the pH value</u>
E cell= 0.0592 V × pH
pH = E cell/ 0.0592 V= 0.182V/ 0.0592V
pH=3.07
The pH of the unknown solution is 3.07.
Density equals mas divided by volume. You know the density and mass so use it to solve for the volume.
d= m/v
d= 19.3 g/mL
m= 50g
v=?
Plug it in. 19.3= 50/v
To solve for v you do the opposite of 50 divided by v which is 50 time v.
That cancels v from the right side of the equation. Do the same on the other side(times v)
19.3 * v=50
Now just divide both sides by 19.3 to get v alone.
Answer:
Answers are in the explanation
Explanation:
Ksp of CdF₂ is:
CdF₂(s) ⇄ Cd²⁺(aq) + 2F⁻(aq)
Ksp = 6.44x10⁻³ = [Cd²⁺] [F⁻]²
When an excess of solid is present, the solution is saturated, the molarity of Cd²⁺ is X and F⁻ 2X:
6.44x10⁻³ = [X] [2X]²
6.44x10⁻³ = 4X³
X = 0.1172M
<h3>[F⁻] = 0.2344M</h3><h3 />
Ksp of LiF is:
LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
Ksp = 1.84x10⁻³ = [Li⁺] [F⁻]
When an excess of solid is present, the solution is saturated, the molarity of Li⁺ and F⁻ is XX:
1.84x10⁻³ = [X] [X]
1.84x10⁻³ = X²
X = 0.0429
<h3>[F⁻] = 0.0429M</h3><h3 /><h3>The solution of CdF₂ has the higher fluoride ion concentration</h3>
