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joja [24]
3 years ago
9

Here is the reaction of carbamic acid and ammonia to form an amide and water. There is a scheme of a reversible reaction where c

arbamic acid reacts with ammonia to give water and an unknown compound. Carbamic acid is H2NCOH with an oxygen atom attached to the carbon atom by a double bond. Ammonia is a nitrogen atom with three H atoms attached. Water is HOH. Draw the amide product of this reaction.
Chemistry
1 answer:
vladimir2022 [97]3 years ago
4 0

Answer:

See explanation

Explanation:

I have attempted to show the sequence of the reaction between carbamic acid and ammonia to form an amide and water and urea.

The reaction first involves the protonation of ammonia to give ammonium carbamate.

When ammonium carbamate is heated to 130-140 degrees, we obtain urea and water as the final products of the reaction

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The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism
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Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

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(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

<em>Where K' = K1 * K2</em>

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just <em>mechanism A and B are consistent with observed rate law</em>

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just <em>mechanism A is consistent with the observation of J. H. Sullivan</em>

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3 0
3 years ago
Help fast !
jok3333 [9.3K]

Answer:

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Given 127.0g of iodine.

no. of iodine atom = 1 mole of iodine

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Given 48g of Mg = 2×6.023×1023

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1 mole of chlorine atom= 6.023× 1023

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3 years ago
The following combinations are not allowed. If n and ml are correct, change the l value to create an allowable combination:
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The allowable combination for the atomic orbital is n=3, l=1, m_{l}=-1.

<h3>What are the three quantum numbers of an atomic orbital?</h3>

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