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Taya2010 [7]
3 years ago
6

Elaborate on the role of activation energy in chemical reactions. A) Decreasing the activation energy leads to higher randomness

within the reaction and an increased reaction rate. B) Activation energy is the minimum energy colliding reactants need to combine and for the chemical reaction to occur. C) The activation energy and randomness indicate the reduced likelihood that the barrier is overcome and a decreased rate. D) Higher activation energy indicates a higher randomness and the energy barrier is overcome with an increased reaction rate.
Chemistry
2 answers:
Zina [86]3 years ago
4 0

<em>B</em><em> </em><em>i</em><em>s</em><em> </em><em>r</em><em>i</em><em>g</em><em>h</em><em>t</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em> </em><em>i</em><em>t</em><em>h</em><em>i</em><em>n</em><em>k</em><em> </em><em>b</em><em>r</em><em>o</em><em>/</em><em>s</em><em>i</em><em>s</em>

Maslowich3 years ago
3 0

Answer:

A and B

Explanation:

The activation energy is the minimum energy necessary for the reaction to happen. It is a barrier, the reaction only occurs if the reactants have that energy.

As higher is the activation energy, as difficult is to the reaction happens, and because of that, the reaction rate will be small. The rate indicates the reaction velocity, which is the velocity of the collisions of the molecules, with more collisions, higher randomness is presented, and higher is the velocity.

A) Correct. As less energy is necessary, as higher is the velocity of the reaction, so the randomness and the rate will increase.

B) Correct. Is the definition of activation energy.

C) Incorrect. They indicate the rate of the reaction, and how easy or not it will be to the reaction happen.

D) Incorrect. As higher is the activation energy, as low is the randomness and the rate of the reaction.

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How does a multicellular organism develop?
crimeas [40]

Answer:

The cells differentiate.

Explanation:

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Multicellular organism contain different types of cells.  The different cells differentiate to perform a particular function. The division of labor property is well shown by the multi cellular organism. The cells of the multi cellular organism must show differentiation process.

Thus, the correct answer is option (d).

7 0
3 years ago
Heterotrophs are the consumers in the food chain, particularly the herbivores, carnivores, and omnivores. A. True B. False
scoundrel [369]
The answer is A. True
5 0
3 years ago
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When the temperature of a sample of neon gas increases, what else increases? Distance between the atoms Kinetic energy of the at
Allushta [10]

Answer: Kinetic Energy of the atoms also increases.

Explanation: We are given that the temperature of the gas increases.

Relation between kinetic energy and temperature follows:

K.E._({avg})=\frac{3RT}{2N_A}

where, K = Average Kinetic energy

R = Gas constant

T = Temperature

N_A = Avogadro's number

As seen from the relation above, the Kinetic energy of the gas is directly proportional to the temperature, hence as the temperature increases, kinetic energy of the atom also increases.

4 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
Which of the following is not a physical change?
lisabon 2012 [21]

Answer:

b or d

Explanation:

4 0
3 years ago
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