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natulia [17]
3 years ago
6

What is the effect of high activation energy on a chemical reaction?

Chemistry
1 answer:
alex41 [277]3 years ago
4 0
<h3><u>Answer;</u></h3>

It makes the reaction harder to start

<h3><u>Explanation</u>;</h3>
  • <em><u>Activation energy is minimum amount of energy that is required for a reaction to start. Activation energy determines the rate of a chemical reaction such that the higher the activation energy, the lower the rate of chemical reaction and vice versa.</u></em>
  • The source of activation energy needed to push chemical reactions forward is obtained from the surroundings. Catalyst speed up chemical reaction by lowering the activation energy. Therefore, catalysis is the increase in the rate of a chemical reaction by lowering its activation energy.
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Dehydration of 2-methyl-2-pentanol forms one major and one minor organic product. Draw the structures of the two organic product
Korvikt [17]

Answer:

The major product is 2-methyl-2-pentene [ CH₃-CH₂-CH=C(CH₃)₂ ] and a minor product 2-methyl-1-pentene [ CH₃-CH₂-CH₂-C(CH₃)=CH₂ ].

Explanation:

Dehydration reaction is a reaction in which a molecule loses a water molecule in the presence of a dehydrating agent like sulfuric acid (H₂SO₄).

<u>Dehydration reaction of 2-methyl-2-pentanol</u> gives a major product 2-methyl-2-pentene and a minor product 2-methyl-1-pentene.

CH₃-CH₂-CH₂-C(CH₃)₂-OH (2-methyl-2-pentanol)→ CH₃-CH₂-CH=C(CH₃)₂ (2-methyl-2-pentene, major) + CH₃-CH₂-CH₂-C(CH₃)=CH₂ (2-methyl-1-pentene, minor)

<u>Since more substituted alkene is more stable than the less substituted alkene. So, the trisubstituted alkene, 2-methyl-2-pentene is more stable than the disubstituted alkene, 2-methyl-1-pentene.</u>

<u>Therefore, the trisubstituted alkene, 2-methyl-2-pentene is the major product and the disubstituted alkene, 2-methyl-1-pentene is the minor product.</u>

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=I_%7B2%7D" id="TexFormula1" title="I_{2}" alt="I_{2}" align="absmiddle" class="latex-formula">
gogolik [260]

Answer: I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

Explanation:

An Oxidant is any substance that oxidizes, or receives electrons from, another; in so doing, it becomes reduced in oxidation number.

A Reductant thus exactly the opposite.

Note that the equation provided shows that Iodine (I2) received an electron to become NEGATIVELY CHARGED:

I2 --> 2I-.

The oxidation number reduced from 0 to -1.

In contrast, the oxidation number of 2S2O3(-2) increases from -4 to -2.

Thus, I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

7 0
3 years ago
One atom of silicon can properly be combined in a compound with
aev [14]
B. two atoms of oxygen
3 0
3 years ago
A "half-reaction” is one which may:
DaniilM [7]

Answer:

d

Explanation:

8 0
3 years ago
Calculate Ecell at 80 ºC for a voltaic cell based on the following redox reaction: H2(g, 1.25 atm) + 2AgCl(s) → 2Ag(s) + 2H+(aq,
iris [78.8K]

Answer:

Ecell = +0.25V

Explanation:

the half-cell reactions for a voltanic cell

cathode(reduction): 2H⁺(aq) + 2e⁻ ------- H₂(g)

anode(oxidation): 2AgCl(s) ------- 2Ag⁺(aq) + 2Cl⁻ + 2e⁻

we have the standard cell potential E⁺cell = 0.18V at 80C respectively

Q = [H⁺]/[Cl⁻]

sub for [H+] = 0.10M and [Cl-] = 1.5M

Q= 0.1M/1.5M

Q = 0.067

Ecell = E⁺cell - \frac{0.059}{n} logQ

= 0.18 - \frac{0.056}{1} log 0.067

0.18- 0.059(-1.174)

Ecell = +0.25V

6 0
3 years ago
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