Answer:
final pressure ( P2) = 467.37 mm Hg
Explanation:
ideal gas:
∴ P1 = 570 mm Hg * ( atm / 760 mm Hg ) = 0.75 atm
∴ T1 = 25 ° C = 298 K
∴ V1 = 1.250 L
∴ R = 0.082 atm L / K mol
⇒ n = P1*V1 / R*T1
⇒ n = (( 0.75 ) * ( 1.25 )) / (( 0.082 ) * ( 298 ))
⇒ n = 0.038 mol gas
∴ T2 = 175 °C ( 448 K )
∴ V2 = 2.270 L
⇒ P2 = nRT2 / V2
⇒ P2 = (( 0.038 ) * ( 0.082 ) * ( 448 )) / 2.270
⇒ P2 = 0.615 atm * ( 760 mm Hg / atm ) = 467.37 mm Hg
It's a combination of factors:
Less electrons paired in the same orbital
More electrons with parallel spins in separate orbitals
Pertinent valence orbitals NOT close enough in energy for electron pairing to be stabilized enough by large orbital size
DISCLAIMER: Long answer, but it's a complicated issue, so... :)
A lot of people want to say that it's because a "half-filled subshell" increases stability, which is a reason, but not necessarily the only reason. However, for chromium, it's the significant reason.
It's also worth mentioning that these reasons are after-the-fact; chromium doesn't know the reasons we come up with; the reasons just have to be, well, reasonable.
The reasons I can think of are:
Minimization of coulombic repulsion energy
Maximization of exchange energy
Lack of significant reduction of pairing energy overall in comparison to an atom with larger occupied orbitals
COULOMBIC REPULSION ENERGY
Coulombic repulsion energy is the increased energy due to opposite-spin electron pairing, in a context where there are only two electrons of nearly-degenerate energies.
So, for example...
↑
↓
−−−−−
↑
↓
−−−−−
↑
↓
−−−−− is higher in energy than
↑
↓
−−−−−
↓
↑
−−−−−
↑
↓
−−−−−
To make it easier on us, we can crudely "measure" the repulsion energy with the symbol
Π
c
. We'd just say that for every electron pair in the same orbital, it adds one
Π
c
unit of destabilization.
When you have something like this with parallel electron spins...
↑
↓
−−−−−
↑
↓
−−−−−
↑
↓
−−−−−
It becomes important to incorporate the exchange energy.
EXCHANGE ENERGY
Exchange energy is the reduction in energy due to the number of parallel-spin electron pairs in different orbitals.
It's a quantum mechanical argument where the parallel-spin electrons can exchange with each other due to their indistinguishability (you can't tell for sure if it's electron 1 that's in orbital 1, or electron 2 that's in orbital 1, etc), reducing the energy of the configuration.
For example...
↑
↓
−−−−−
↑
↓
−−−−−
↑
↓
−−−−− is lower in energy than
↑
↓
−−−−−
↓
↑
−−−−−
↑
↓
−−−−−
To make it easier for us, a crude way to "measure" exchange energy is to say that it's equal to
Π
e
for each pair that can exchange.
So for the first configuration above, it would be stabilized by
Π
e
(
1
↔
2
), but the second configuration would have a
0
Π
e
stabilization (opposite spins; can't exchange).
PAIRING ENERGY
Pairing energy is just the combination of both the repulsion and exchange energy. We call it
Π
, so:
Π
=
Π
c
+
Π
e
Inorganic Chemistry, Miessler et al.
Inorganic Chemistry, Miessler et al.
Basically, the pairing energy is:
higher when repulsion energy is high (i.e. many electrons paired), meaning pairing is unfavorable
lower when exchange energy is high (i.e. many electrons parallel and unpaired), meaning pairing is favorable
So, when it comes to putting it together for chromium... (
4
s
and
3
d
orbitals)
↑
↓
−−−−−
↑
↓
−−−−−
↑
↓
−−−−−
↑
↓
−−−−−
↑
↓
−−−−−
↑
↓
−−−−−
compared to
↑
↓
−−−−−
↑
↓
−−−−−
↑
↓
−−−−−
↑
↓
−−−−−
↑
↓
−−−−−
↑
↓
−−−−−
is more stable.
For simplicity, if we assume the
4
s
and
3
d
electrons aren't close enough in energy to be considered "nearly-degenerate":
The first configuration has
Π
=
10
Π
e
.
(Exchanges:
1
↔
2
,
1
↔
3
,
1
↔
4
,
1
↔
5
,
2
↔
3
,
2
↔
4
,
2
↔
5
,
3
↔
4
,
3
↔
5
,
4
↔
5
)
The second configuration has
Π
=
Π
c
+
6
Π
e
.
(Exchanges:
1
↔
2
,
1
↔
3
,
1
↔
4
,
2
↔
3
,
2
↔
4
,
3
↔
4
)
Technically, they are about
3.29 eV
apart (Appendix B.9), which means it takes about
3.29 V
to transfer a single electron from the
3
d
up to the
4
s
.
We could also say that since the
3
d
orbitals are lower in energy, transferring one electron to a lower-energy orbital is helpful anyways from a less quantitative perspective.
COMPLICATIONS DUE TO ORBITAL SIZE
Note that for example,
W
has a configuration of
[
X
e
]
5
d
4
6
s
2
, which seems to contradict the reasoning we had for
Cr
, since the pairing occurred in the higher-energy orbital.
But, we should also recognize that
5
d
orbitals are larger than
3
d
orbitals, which means the electron density can be more spread out for
W
than for
Cr
, thus reducing the pairing energy
Π
.
That is,
Π
W
<span>Starch and
cellulose have the same substance but different structures. They are both
polysaccharides. The basic unit of a polysaccharide is the glucose. Glucose,
which contains carbon, hydrogen, and oxygen, have two forms. The alpha-glucose
with an alcohol group attached to carbon 1 is down and the beta-glucose with
the alcohol group attached to carbon 1 is up. Starch is the alpha-glucose while
cellulose is the beta-glucose. Starches are linked into a straight chain whereas
the cellulose are connected like a pile of stack paper. When the human body
eats starch, it can digest the starch but not the cellulose because it has no
enzyme that can break it down. </span>
None because Covalent compounds don’t conduct electricity because they are formed between the non metal atoms by sharing of electrons. The Covalent compounds haves no free electrons and also no ions and hence they do not conduct electricity. That is why they do not conduct electricity.
Answer:
C）52g KCl in 100g water at 80°C
Explanation:
A saturated solution is one that contains as much solute as it can dissolve in the presence of excess solute at that particular temperature.
A solutibility curve is a graph that shows the variability with temperature of the solubility of a solute in a given solvent. A solutibility curve can provide information of whether a solution formed frommthe solute and solvent are saturated or not at a given temperature.
From the solubility curve in the attachment below:
A) A saturated solution of NH₄Cl will contain about 52 g solute per 100 g sat 50 °C. Thus, a solution of 40 g NH₄Cl in 100 g water at 50 °C is an unsaturated solution.
B) A saturated solution of SO₂ at 10°C will contain about 70 g of solute in 100 g of water. Thus a solution of 2g SO₂ in 100g water at 10°C is an unsaturated solution.
C) A saturated solution of KCl at 80 °C will contain about 52 g of solute in 100 g of water. Thus, a solution of 52g KCl in 100g water at 80°C is a saturated solution.
D) A saturated solution of Kl at 20 °C will contain about 145 g of solute in 100 g of water. Thus, a solution of 120g KI in 100g water at 20°C is an unsaturated solution.
