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joja [24]
2 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]2 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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san4es73 [151]

Answer:

B oklolol

[tex]\purple{\rule{45pt}{7pt}}\purple{\rule{45pt}{999999pt}}[tex]

8 0
2 years ago
7.0 mol Mn reacts with 5.0 mol
Elenna [48]

The moles of MnO formed with the reaction of 5 moles of Mn is 3 moles.

<h3>What is a limiting reagent?</h3><h3 />

In a chemical reaction, the reagent that is present in a lesser quantity and governs the rate of the reaction is termed as the limiting reagent.

In the reaction of the formation of MnO, according to the stoichiometric law 2 moles of Mn reacts with 1 moles of Oxygen.

Thus, the moles of oxygen consumed by 7 moles of Mn is:

2 moles Mn = 1 mole O₂

7 moles Mn = 3.5 moles O₂

The available moles of O₂ = 5 mol.

The remaining moles of O₂ = 5-3.5 moles

The remaining moles of O₂ = 1.5 moles

The reaction of 5 moles Mn requires 2.5 moles of O₂, whereas the available moles of oxygen is 1.5 moles. Thus, oxygen serves as the limiting reagent.

The moles of MnO formed with the reaction of 5 moles Mn and 1.5 moles O₂

1 mole O₂ = 2 moles MnO

1.5 moles O₂ = 2 * 1.5 moles MnO

1.5 moles O₂ = 3 moles MnO

Thus, the moles of MnO formed with the reaction of 5 moles of Mn with the available oxygen is 3 moles.

Learn more about stoichiometric law, here:

brainly.com/question/14465605

#SPJ1

6 0
1 year ago
Read 2 more answers
Calculate the density a rectangular block of a metal whose length is 8.335cm width is 1.02cm height is 0.982cm and mass is 62.35
Ostrovityanka [42]

Answer:

Density rectangular block = 7.47 (Approx) gm/cm³

Explanation:

Given:

Length = 8.335 cm

Width = 1.02 cm

Height = 0.982 cm

Mass = 62.3538 gm

Find:

Density rectangular block

Computation:

Volume of block = lbh

Volume of block = (8.335)(1.02)(0.982)

Volume of block = 8.3486 cm³

Density = Mass / Volume

Density rectangular block = 62.3538 / 8.3486

Density rectangular block = 7.47 (Approx) gm/cm³

8 0
2 years ago
A 3.81-gram sample of NaHCO3 was completely decomposed in an experiment.
givi [52]

Explanation:

In this experiment, carbon dioxide and water vapors combine to form H2CO3. After decomposition, the Na2CO3 had a mass of 2.86 grams. Determine ...

8 0
3 years ago
Compute 4.659×104−2.14×104. Round the answer appropriately.
antiseptic1488 [7]
<span>The answer is 2.519 × 10^4. Use the distributive property: a × x + b × x = (a + b) × x. If a = 4.659, b = 2.14, and x = 10^4, then 4.659 × 10^4 − 2.14 × 10^4 = (4.659 - 2.14) × 10^4. Now, subtract numbers in parenthesis: 4.659 × 10^4 − 2.14 × 10^4 = (4.659 - 2.14) × 10^4 = 2.519 × 10^4.Hope this helps. Let me know if you need additional help!</span>
7 0
3 years ago
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