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3 years ago

Acetone major species present when dissolved in water

Chemistry

1 answer:

jek_recluse [69]3 years ago

3 0

Answer: acetone molecule ( CH₃-CO-CH₃)

Explanation:

1) Acetone is CH₃-CO-CH₃

2) That is a molecule (build up of covalent bonds).

3) When dissolved, covalent bonded compounds remain as separate molecules, then it is said that the major species present in the solution is the molecule. The molecules of acetone are surrounded (sovated) by the molecules of water.

This as opposed to the case of ionic compounds that ionize. When a compound as NaCl dissolves in water, it ionizes completely, so the major speceies are not NaCl formulas, but the ions Na⁺ and Cl⁻, not molecules.

That leads to the answer: the major species present when acetone is dissolved in water is the molecules of acetone (you do not need to state the fact that the molecules of water are part of the solution, because that is not the target of the question).

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Calculate the wavelength, in nanometers, of the light emitted by a hydrogen atom when its electron falls from the n = 7 to the n

Gre4nikov [31]

Answer:

The wavelength of the light emitted by a hydrogen atom for the given transition is 2166 nm.

Explanation:

The energy of nth energy levels of the H atom is given as:

$E_n = -2.18 \times 10^{-18} \times \frac{1}{n^2} J$

Energy of the seventh energy level = $E_7$

$E_7=-2.18 \times 10^{-18} \times \frac{1}{7^2} J$

$E_7=-2.18 \times 10^{-18} \times \frac{1}{7^2} J=-4.4490\times 10^{-20} J$

Energy of the seventh energy level = $E_4$

$E_4=-2.18 \times 10^{-18} \times \frac{1}{4^2} J$

$E_4=-2.18 \times 10^{-18} \times \frac{1}{16} J=-1.3625\times 10^{-19} J$

Energy of the light emitted will be equal to the energy difference of the both levels.

$E=E_7-E_4=-4.4490\times 10^{-20} J-(-1.3625\times 10^{-19} J)$

$E=9.176\times 10^{-20} J$

Wavelength corresponding to energy E can be calculated by using Planck's equation:

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

$\lambda =\frac{hc}{E}=\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{9.176\times 10^{-20} J}=2.166\times 10^{-6} m=2166 nm$

The wavelength of the light emitted by a hydrogen atom for the given transition is 2166 nm.

8 0

3 years ago

Can someone help me please:)

docker41 [41]

Answer:

what we ain't that smart

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3 years ago

A decrease in height of a column in a mercury barometer means that

insens350 [35]

Answer:

increase in the air pressure

Explanation:

The barometer is a device that is used for measuring the air pressure. It is a device that uses mercury in order to show the air pressure. The mercury reacts easily to the changes in the air pressure, so it is a nice indicator for it. The air pressure can simply be defined as the weight of the air masses, and the pressure they make it because of it on the objects. The lower the air pressure, the higher up the mercury will go, as it will experience less pressure from the air, and the higher the air pressure, the lower the mercury will drop, as it will experience more pressure from the air.

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3 years ago

When an unknown amine reacts with an unknown acid chloride, an amide with a molecular mass of 163 g/mol (M = 163 m/z) is formed.

DanielleElmas [232]

Answer:

As the molecular mass of given amine is 163 g/mol (a odd number) it means that this compound contains a odd number of Nitrogen atoms. We will first apply Rule of Thirteen to get the molecular formula.

Rule of Thirteen:

First divide the parent peak value by 13 as,

= 163 ÷ 13

= 12.53

Now, multiply 13 by 12,

= 13 × 12 (here, 12 specifies number of carbon atoms)

= 156

Now subtract 156 from 163,

= 163 - 156

= 7

Add 7 into 12,

= 7 + 12

= 19 (hydrogen atoms)

So, the rough formula we have is,

C₁₂H₁₉

Now, add one Nitrogen atom to above formula and subtract one Carbon and 2 Hydrogen atoms as these numbers are equal to atomic mass of Nitrogen atom as,

C₁₂H₁₉ -------N--------> C₁₁H₁₇N

Also, as shown in ¹³C-NMR there is one peak around 180 ppm and the peak at 1661 cm⁻¹ in IR spectrum is characteristic to carbonyl group hence, we will add one oxygen atom to the chemical formula accordingly. i.e.

C₁₁H₁₇N -------O--------> C₁₀H₁₃NO

Molecular Formula: C₁₀H₁₃NO

Also,

In NMR the the four peaks around 120 ppm are assigned to a mono substituted benzene ring.

The absence of IR peak above 3200 cm⁻³ also confirms that the amine is tertiary in nature and there is no hydrogen attached to the nitrogen atom.

It can be observed that the peaks in upfield are duplicating. This can be due to the presence of rotamers of said compound.

The most plausible structure for given data is shown below, and the resonance structure along with rotamers are also shown.

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viktelen [127]

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