The conversion of acetyl chloride to methyl acetate occurs via the following two-step mechanism: Reaction sequence of acetyl chl
1 answer:
If the concentration of acetyl chloride is increased ten times the rate of reaction is increased ten times.
The conversion of acetyl chloride to methyl acetate is a substitution reaction. Recall that a substitution reaction is one in which a moiety in a molecule is replaced by another.
In this reaction, the CH3O- ion replaces the chloride ion. In the first step, the CH3O- ion attacks the substrate in a slow step. This creates a tetrahedral intermediate. Loss of the chloride ion yields the methyl acetate product.
The rate determining step is the formation of the tetrahedral intermediate. Since the reaction is first order in the acetyl chloride, if its concentration is increased ten times the rate of reaction is increased ten times.
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Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is .
Explanation:
When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.
For example, chemical equation for oxidation of methane is as follows.
Number of atoms present on reactant side are as follows.
- C = 1
- H = 4
- O = 2
Number of atoms present on product side are as follows.
- C = 1
- H = 2
- O = 3
To balance this equation, multiply by 2 on reactant side. Also, multiply
by 2 on product side. Hence, the equation can be rewritten as follows.
Now, the number of atoms present on reactant side are as follows.
- C = 1
- H = 4
- O = 4
Number of atoms present on product side are as follows.
- C = 1
- H = 4
- O = 4
Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.
Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is .
Answer:
28g remain after 13.5 hours
Explanation:
Element decayment follows first order kinetics law:
ln[Pa-234] = -kt + ln [Pa-234]₀ <em>(1)</em>
<em>Where [Pa-234] is concentration after t time, k is rate constant in time, and [Pa-234]₀ is initial concentration</em>
Half-life formula is:
6.75 = ln2 / k
<em>k = 0.1027hours⁻¹</em>
Using rate constant in (1):
ln[Pa-234] = -0.1027hours⁻¹×13.5hours + ln [112.0g]
ln[Pa-234] = 3.332
[Pa-234] = <em>28g after 13.5 hours</em>
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