Question

1g of urea (NH2CONH2),1g acetic acid (CH3COOH)and 1g of formaldehyde (HCHO)will have H- atom is in the ratio:

I want the answer with very detailed(step by step) expalnation.

Answer 1

The correct question may be like:

"If we have 1 gram of urea (NH2CONH2), 1 gram of acetic acid (CH3COOH), and 1 gram of formaldehyde (HCHO), what will be the ratio of H-atoms in these compounds?

To find the ratio of H-atoms, we need to first determine the number of H-atoms in each compound.

Urea (NH2CONH2) has 4 H-atoms. This is because there are two NH2 groups in the molecule, and each NH2 group has two H-atoms.

Acetic acid (CH3COOH) has 4 H-atoms. This is because there are two CH3 groups in the molecule, and each CH3 group has three H-atoms. Additionally, there is one H-atom in the COOH group.

Formaldehyde (HCHO) has 2 H-atoms. This is because there is one H-atom attached to the carbon atom, and one H-atom attached to the oxygen atom.

Now that we know the number of H-atoms in each compound, we can find the ratio of H-atoms by adding up the total number of H-atoms and dividing each compound's number of H-atoms by the total.

Total number of H-atoms = 4 + 4 + 2 = 10

Urea: 4/10 = 0.4

Acetic acid: 4/10 = 0.4

Formaldehyde: 2/10 = 0.2

Therefore, the ratio of H-atoms in these compounds is 0.4:0.4:0.2, or simplified, 2:2:1."

The ratio of H-atoms in the given compounds is 2:2:1. To arrive at this answer, we first determined the number of H-atoms in each compound. Urea has 4 H-atoms, acetic acid has 4 H-atoms, and formaldehyde has 2 H-atoms. We then added up the total number of H-atoms, which was 10. Finally, we divided each compound's number of H-atoms by the total and simplified the ratio to 2:2:1.

For such more question on formaldehyde

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