Chemistry College

1. Write the IUPAC names for the following 1.1 1.2 N 1.3 O NO2 x Y ·0 OH 5​

Answers

Answer 1

1. The IUPAC name of N is nitrogen.

2. Nitrogen dioxide

3.The IUPAC name of O is oxygen

4.The IUPAC name of OH is hydroxyl.

The IUPAC name of ·0 is a radical. It is commonly found in organic chemistry and plays an important role in many reactions.

IUPAC names for the given compounds are:1.1. N: Nitrogen

The IUPAC name of N is nitrogen.

It is a non-metal and belongs to group 15 in the periodic table. It has an electronic configuration of 1s2 2s2 2p3.1.2. NO2: Nitrogen dioxide

Explanation: NO2 is a chemical compound that is formed by the combination of nitrogen and oxygen. It is a reddish-brown gas that has a pungent odor.

The IUPAC name of NO2 is nitrogen dioxide.1.3. O: Oxygen

Explanation: The IUPAC name of O is oxygen.

It is a non-metal and belongs to group 16 in the periodic table. It has an electronic configuration of 1s2 2s2 2p4.

X: UnknownExplanation: No IUPAC name can be given to an unknown compound as the structure and composition are not known.

Y: Hydroxyl Explanation: The IUPAC name of OH is hydroxyl.

It is a functional group that is composed of an oxygen atom and a hydrogen atom (-OH). It is commonly found in alcohols and phenols. ·0: RadicalExplanation: A radical is a molecule or an ion that contains an unpaired electron.

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Note: The complete question is given below

Provide the IUPAC names for the following compounds:

[tex]CH_3CH_2CH(CH_3)CH_2CH_2CH_2CH_3[/tex]

C6H5CH(CH3)2

H2NCH2CH2CH2CH2CH2NH2

CH3CH2CH2CH2CH2OH

CH3CH2CH2CHOHCH3

Related Questions

Drag the tiles to the correct locations on the equation. Not all tiles will be used.
Two atoms interact with each other and change as shown by the equation. Complete the equation by filling in the missing parts.
5
2
4
3
1
H+H -
H
He
Li
+

Answers

The equation  in the question is: H+H → H + H Complete the equation by filling in the missing parts. missing part is  1 → H+H-2 → →3 → He.

The atomic number of hydrogen is 1, which means it has only one proton in the nucleus and one electron in its shell. Two hydrogen atoms react with each other to form helium. Helium has 2 protons and 2 neutrons in its nucleus and two electrons in its shell. Therefore, the equation is:

H + H → HeIt can be seen that:1. H + H (Reactants)

2. → (Yields or Reacts to form)

3. He (Product)Therefore, the tiles will be arranged as shown below: 1 → H+H-2 → →3 → He

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In using the Haber process in the formation of ammonia, what mass of hydrogen is needed to produce 51.0 grams of ammonia? 3 H₂(g) + N2 (g) → 2 NH3(g).
​

Answers

The mass of hydrogen needed to produce 51.0 grams of ammonia is  ≈ 9.07 grams.

To determine the mass of hydrogen required to produce 51.0 grams of ammonia (NH3) using the Haber process, we need to calculate the stoichiometric ratio between hydrogen and ammonia.

From the balanced chemical equation:

3 H₂(g) + N₂(g) → 2 NH₃(g)

We can see that for every 3 moles of hydrogen (H₂), we obtain 2 moles of ammonia (NH₃).

First, we need to convert the given mass of ammonia (51.0 grams) to moles. The molar mass of NH₃ is 17.03 g/mol.

Number of moles of NH₃ = Mass / Molar mass

                     = 51.0 g / 17.03 g/mol

                     ≈ 2.995 moles

Next, using the stoichiometric ratio, we can calculate the moles of hydrogen required.

Moles of H₂ = (Moles of NH₃ × Coefficient of H₂) / Coefficient of NH₃

           = (2.995 moles × 3) / 2

           ≈ 4.493 moles

Finally, we can convert the moles of hydrogen to mass using the molar mass of hydrogen (2.02 g/mol).

Mass of H₂ = Moles × Molar mass

          = 4.493 moles × 2.02 g/mol

          ≈ 9.07 grams

Therefore, approximately 9.07 grams of hydrogen is needed to produce 51.0 grams of ammonia in the Haber process.

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In using the Haber process in the formation of ammonia, what mass of hydrogen is needed to produce 51.0 grams of ammonia? 3 H₂(g) + N2 (g) → 2 NH3(g).
​

Answers

To produce 51.0 grams of ammonia using the Haber process, approximately 76.5 grams of hydrogen is needed (based on the stoichiometry of the balanced equation).

In the given balanced equation for the Haber process, it states that three moles of hydrogen gas (H₂) react with one mole of nitrogen gas (N₂) to produce two moles of ammonia gas (NH₃).

To determine the mass of hydrogen needed to produce 51.0 grams of ammonia, we need to set up a proportion using the molar masses and stoichiometric coefficients from the balanced equation.

The molar mass of hydrogen is approximately 2.02 g/mol, and the molar mass of ammonia is approximately 17.03 g/mol.

Using the proportion:

(3 mol H₂ / 2 mol NH₃) = (x g H₂ / 51.0 g NH₃)

Cross-multiplying and solving for x (mass of hydrogen), we get:

x g H₂ = (3 mol H₂ / 2 mol NH₃) * (51.0 g NH₃)

x g H₂ = 76.5 g H₂

Therefore, to produce 51.0 grams of ammonia, approximately 76.5 grams of hydrogen is needed in the Haber process.

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