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Serggg [28]
2 years ago
13

Why it is that evaporation causes cooling,but boiling causes heating?

Chemistry
2 answers:
musickatia [10]2 years ago
8 0
During evaporation, particles which have a high kinetic energy come out of the body by gaining heat due to which there is a cooling effect on the particle of the body itself. that is why, evaporation causes cooling. For example: sweating 
rusak2 [61]2 years ago
7 0
In evaporation due to internal heat, kinetic energy of molecules increases and they come to the top and take out that heat with them when they evaporate thus causes cooling 
in boiling as heat is given to the molecules so their kinetic energy increases and they start vibrating with great energy and thus causes heating
You might be interested in
Please help, please don’t answer links
lakkis [162]

Answer:

0.093 mole of C₆H₁₂.

Explanation:

We'll begin by calculating the molar mass of C₆H₁₂. This can be obtained as follow:

Molar mass of C₆H₁₂ = (12×6) + (12×1)

= 72 + 12

= 84 g/mol

Finally, we shall determine the number of mole in 7.8 g of C₆H₁₂. This can be obtained as follow:

Molar mass of C₆H₁₂ = 84 g/mol

Mass of C₆H₁₂ = 7.8 g

Mole of C₆H₁₂ =?

Mole = mass / molar mass

Mole of C₆H₁₂ = 7.8 / 84

Mole of C₆H₁₂ = 0.093 mole

Thus, 7.8 g contains 0.093 mole of C₆H₁₂.

3 0
2 years ago
1. If 500 ML of a 3 L sample of 0.20 M sodium chloride solution is spilled, what is the concentration of the remaining solution?
photoshop1234 [79]
D. 0.2 M

The concentration of a solution is basically the ratio of the solute present to the solvent in the solution. This is an intrinsic property, independent of the amount of solution that is present. A similar example is that of density. No matter the size of a sample, the density and concentration of that sample remain constant.
3 0
3 years ago
Calculate the molar solubility of fe(oh)2 in pure water. (the value of ksp for fe(oh)2 is 4.87×10−17.)
rosijanka [135]
When we have this balanced equation for a reaction:
Fe(OH)2(s) ↔ Fe+2  + 2OH-

when Fe(OH)2 give 1 mole of Fe+2 & 2 mol of OH-
so we can assume [Fe+2] = X and [OH-] = 2 X
 when Ksp = [Fe+2][OH-]^2
and have Ksp = 4.87x10^-17 
[Fe+2]= X
[OH-] = 2X
so by substitution 
4.87x10^-17 = X*(2X)^2
∴X^3 = 4.8x10^-17 / 4
∴the molar solubility X = 2.3x10^-6 M
3 0
3 years ago
1s^2 2s^2 2p^6 3s^2 3p^6 how many unpaired electrons are in the atom represented by the electron configuration above?
Sedbober [7]
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
5 0
2 years ago
Read 2 more answers
What is the complement DNA strand to GTACCTAGGT?
Serga [27]
CATGGATCCA for future reference this is considered biology, you may get a quicker response if you categorize it as such :)
6 0
2 years ago
Read 2 more answers
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