Rutherford’s experimental evidence led to the development of this atomic model from the one before it . According to him , a few of the positive particles pointed at a gold foil appeared to ricochet back off of the light metallic foil. This experimental evidence gives the root the development of the atomic model .
Answer:
Mass of proton is
<h2>
1.6726219 × 10-27 Kg</h2>
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Description in biology, cell theory is the historic scientific theory, now universally accepted, that living organisms are made up of cells, that they are the basic structural/organizational unit of all organisms, and that all cells come from pre-existing cells.
Answer:
a) LAW OF MULTIPLE PROPORTIONS
b) 0.095g, 0.71g, 0.285g respectively
c) X2Y, YZ15, X6Y
d) hence mass of compound X = 21 x 0.045 = 0.95g
mass of compound Y = 21 x 0.955 = 20.05g
Explanation:
a) The assumptions made in solving this questions is the application of the LAW OF MULTIPLE PROPORTIONS. The Law of multiple proportions states that if two elements A and B combine together to form more than one compound, then the several masses of A which chemically combine with a fixed mass of B is in a simple ratio.
for example, copper forms two oxides ; copper(I) oxide (CuO) and copper(ii) oxide(Cu2O), it is possible for the two samples of the oxides to be reduced to Cu by reacting with Hydrogen gas. as such, certain masses of oxygen combine separately with a fixed mass of Cu. then the ratios of Cu are then determined.
b) To calculate the relative masses, we take note of the three compounds given, they all have some amount of Y in them, hence we can use Y as our relative mass, this implies that the relative mass of Y = 1g
mass of X = 0.4g
mass of Y = 4.2g
amount of X in 1g of Y = 0.4 x 1 /4.2
= 0.095g
for compound 2;
mass of Y = 1.4g
mass of Z = 1.0g
amount of Z in 1g of Y =1.0 x 1 /1.4
= 0.71g
for compound 3;
mass of X = 2.0g
mass of Y = 7.0g
amount of X in 1g of Y = 1 x 2/7
= 0.285g
c) Applying the law of multiple proportions; since elements X and Z combine with a fixed mass of Y, they must bear a simple ratio;
compound 1/compound 3 = 0.095/0.285
= 1/3
compound 1/compound 2 = 0.095/0.71
= 2/15
compound 2/ compound 3 = 0.71/0.285
= 5/2
formular for compound 1 = X2Y
formula for compound 2 = YZ15
formular for compound 3 = X6Y
d) from the formular X2Y, we can get the amount of each product in XY using the ratios
%of compound XY in X = mass of compound X / total Mass
= 0.2/4.4 = 4.5%
as such in a 21g of compound XY, %of compound Y = 1 - %of compound X = 95.5%
hence mass of compound X = 21 x 0.045 = 0.95g
mass of compound Y = 21 x 0.955 = 20.05g
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
