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____ [38]
3 years ago
10

Why are aldehydes more reactive than ketones?​

Chemistry
1 answer:
Simora [160]3 years ago
5 0

Answer:

Aldehydes are typically more reactive than ketones due to the following factors. ... The carbonyl carbon in aldehydes generally has more partial positive charge than in ketones due to the electron-donating nature of alkyl groups. Aldehydes only have one e- donor group while ketones have two.

Explanation:

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It’s a nonet because an octet is 8, a quartet is 4, and a duet is 2.
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A book sits motionless on a table. Which statement is true? Group of answer choices There are no forces acting on the book. The
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Answer:

The forces acting on the book are balanced by each other.

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2 years ago
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
What are 4 similarities of Metals, Nonmetals, and Metalloids?
igor_vitrenko [27]
The answer is           


e o i and a



not really sure forgive me if im wrong


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If a mixture containing 0.50 moles of N2 and 1.8 moles of H2 is allowed to react according to the equation N2 3H2 2NH3 Group of
Lelechka [254]

Answer: hydrogen is the limiting reactant.

Explanation:

We have the equation \text{N}_{2}+3\text{H}_{2} \longrightarrow 2\text{NH}_{3}.

This means that for every mole of nitrogen consumed, 3 moles of hydrogen are consumed.

  • Considering the nitrogen, the reaction can occur 0.50 times.
  • Considering the hydrogen, the reaction can occur 1.8/3 = 0.6 times.

Therefore, <u>hydrogen</u> is the limiting reactant.

7 0
2 years ago
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