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2 years ago

13

Why nitro group is called an ambident group?

2 answers:

Zanzabum2 years ago

7 0

Answer:

The nitro group is an ambident group and is capable of getting attached to carbon chain through nitrogen. as well as through oxygen (-O - N = O) atom. The compound in which the -NO2 group is linked to the alkyl or aryl group through oxygen atom are called nitrites

podryga [215]2 years ago

4 0

Answer:

It can bond through either the N or O atoms

Explanation:

An ambident ("both teeth") group is a group that can attach to another by either of two atoms.

A nitro group can bond through either N or O.

Organic chemistry

In organic chemistry, this property gives rise to functional group isomers.

For example, we can have either nitromethane (CH₃NO₂, Fig. 1) or methyl nitrite (CH₃ONO, Fig. 2)

Inorganic chemistry

The ambidentate property of the nitro group gives rise to linkage isomers in inorganic coordination complexes.

For example, we can have either a pentachloronitroiron(III) cation (Fig. 3) or a pentachloronitritoiron(III) cation (Fig. 4)

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Predict the initial and isolated products for the reaction. The starting material is a 6 carbon chain where there is a triple bo

satela [25.4K]

Answer:

See explanation and image attached

Explanation:

This reaction is known as mercuric ion catalyzed hydration of alkynes.

The first step in the reaction is attack of the mercuric ion on the carbon-carbon triple bond, a bridged intermediate is formed. This bridged intermediate is attacked by water molecule to give an organomercury enol. This undergoes keto-enol tautomerism, proton transfer to the keto group yields an oxonium ion, loss of the mercuric ion now gives equilibrium keto and enol forms of the compound. The keto form is favoured over the enol form.

7 0

2 years ago

agasfer [191]

Answer:

3.02× 10²⁴ atoms

Explanation:

Given data:

Number of nitrogen atoms = ?

Number of moles of N₂O = 2.51 mol

Solution:

1 mole contain 2 mole of nitrogen atoms.

2.51 × 2 = 5.02 mol

According to Avogadro number,

1 mole = 6.022 × 10²³ atoms

5.02 mol × 6.022 × 10²³ atoms / 1 mol

30.2 × 10²³ atoms

3.02× 10²⁴ atoms

5 0

1 year ago

What is the name of the following ionic compound?: MoAs

Minchanka [31]

Molybdenum Arsenide

I think that’s right but not %100 sure

3 0

2 years ago

What is the mole fraction of potassium dichromate, K2Cr2O7, in a solution prepared from 24.42 g of potassium dichromate and 240.

Harrizon [31]

Answer:

$6.19\times 10^{-3}$ is the mole fraction of potassium dichromate.

Explanation:

Mass of potassium dichromate = 24.42 g

Moles of potassium dichromate = $n_1=\frac{24.42 g}{294.185 g/mol}=0.0830 mol$

Mass of water = 240.0 g

Moles of water = $n_2=\frac{240.0 g}{18.015 g/mol}=13.3222 mol$

Mole fraction is calculated by:

$\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{0.0830 mol}{0.0830 mol+13.3222 mol}=0.00619=6.19\times 10^{-3}$

$6.19\times 10^{-3}$ is the mole fraction of potassium dichromate.

8 0

2 years ago

One of the emission spectral lines for Be31 has a wavelength of 253.4 nm for an electronic transition that begins in the state w

max2010maxim [7]

Explanation:

The given data is as follows.

$\lambda$ = 253.4 nm = $253.4 \times 10^{-9}m$ (as 1 nm = $10^{-9}$ )

$n_{1}$ = 5, $n_{2}$ = ?

Relation between energy and wavelength is as follows.

E = $\frac{hc}{\lambda}$

= $\frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{253.4 \times 10^{-9}}$

= $0.0784 \times 10^{-17}$ J

= $7.84 \times 10^{-19} J$

Hence, energy released is $7.84 \times 10^{-19} J$ .

Also, we known that change in energy will be as follows.

$\Delta E = -2.178 \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}$

where, Z = atomic number of the given element

$7.84 \times 10^{-19} J = -2.178 \times (4)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

$\frac{7.84 \times 10^{-19} J}{34.848} = \frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

0.02 + 0.04 = $\frac{1}{n^{2}_{1}}$

$n_{1}$ = $\sqrt{\frac{1}{0.06}}$

= 4

Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.

4 0

2 years ago