Answer:
5 moles of NO₂ will remain after the reaction is complete
Explanation:
We state the reaction:
3NO₂(g) + H₂O(l) → 2HNO₃(l) + NO(g)
3 moles of nitric oxide can react with 1 mol of water. Ratio is 3:1, so we make this rule of three:
If 3 moles of nitric oxide need 1 mol of water to react
Then, 26 moles of NO₂ may need (26 .1) / 3 = 8.67 moles of H₂O
We have 7 moles of water but we need 8.67 moles, so water is the limiting reactant because we do not have enough. In conclusion, the oxide is the reagent in excess. We can verify:
1 mol of water needs 3 moles of oxide to react
Therefore, 7 moles of water will need (7 .3)/1 = 21 moles of oxide
We have 26 moles of NO₂ and we need 21, so we still have oxide after the reaction is complete. We will have (26-21) = 5 moles of oxide that remains
Answer : The one diagram which shows the electron with the highest potential energy is attached below.
Explanation : One can easily find the highest potential energy of the atom just by looking at the diagram, the electron which is from farthest distance from the atomic nucleus will have the highest potential energy in the electron.
Answer:
10.8 days (3 sig.figs.)
Explanation:
All radioactive decay is 1st order decay defined by the expression A = A₀e^-kt
which is solved for time of decay (t) => t = ln(A/A₀) / -k
A = final weight = 1.0 gram
A₀ = initial weight = 16.0 grams
k = rate constant = 0.693/t(1/2) = 0.693/2.69 days = 0.258 days⁻¹
t = ln(1/16) / -0.258da⁻¹ = (-2.77/-0.258) days = 10.74646792 days (calculator)
≅ 10 days (1 sig. fig. based on given 1 gram mass)
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
The effective speed (rms) of the oxygen gas is 293.68 m/s.
<h3>
</h3><h3>What is Root-mean-square velocity?</h3>
Root mean square velocity is the square root of the mean of squares of the velocity of individual gas molecules
![v_{rms}=\sqrt[]{\frac{3RT}{M} }](https://tex.z-dn.net/?f=v_%7Brms%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B3RT%7D%7BM%7D%20%7D)
<em>where </em>R = universal gas constant
M = molar mass of the gas in kg/mol
T = temperature in Kelvin
According to the ideal gas law,
PV = nRT
RT = 
Substitute in the rms velocity formula,
![v_{rms} = \sqrt[]{\frac{3PV}{nM} }](https://tex.z-dn.net/?f=v_%7Brms%7D%20%3D%20%5Csqrt%5B%5D%7B%5Cfrac%7B3PV%7D%7BnM%7D%20%7D)
P = 92 kPa, V = 10 L, n = 2 moles and M = 32 x 10⁻³ kg/mol
![v_{rms} = \sqrt[]{\frac{3\times92\times10}{2\times32\times10^-^3} }](https://tex.z-dn.net/?f=v_%7Brms%7D%20%3D%20%5Csqrt%5B%5D%7B%5Cfrac%7B3%5Ctimes92%5Ctimes10%7D%7B2%5Ctimes32%5Ctimes10%5E-%5E3%7D%20%7D)
=293.68 m/s
Thus, the effective speed (rms) of O₂ gas is 293.68 m/s.
Learn more about Root-mean-square velocity:
brainly.com/question/15995507
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