What is largest value for the third quantum number

What should have only one idependent variable

jenyasd209 [6]

I think a controlled experiment

3 0

3 years ago

Substituents on an aromatic ring can have several effects on electrophilic aromatic substitution reactions. Substituents can act

Leya [2.2K]

Answer:

Option A=> -NHCOCH3 and option C = -CH3.

Explanation:

The option A that is -NHCOCH3 is CORRECT because it possesses lone pair of electron with the exception of group 7A elements. It is this lone pair that is used in the Activation of the ring towards substitution. Other groups that falls into this group are; OCH3, alkyls and many others.

Option B that is -COOH is good group for withdrawal of electron through Resonance. Other examples are NO2, -CN and SO3H.

Option C falls to the same category as option A above that is Activation of the ring towards substitution.

Option D falls to the same category as option B above that is group for withdrawal of electron through Resonance.

7 0

3 years ago

Consider the Bohr model of the simplest atom, hydrogen. What energy photon will be released if an electron falls from the n= 2 o

Nana76 [90]

<h2>Ultraviolet Light</h2>

Explanation:

The energy of a photon that will be released if an electron falls from the n= 2 orbit (excited state) to the n0 = 1 orbit (ground state) is of ultraviolet light.
In the ultraviolet part of the spectrum, a photon having an energy of 10.2 eV has a wavelength of 1.21 x 10-7 m.
Hence, when an electron wants to jump or it gets excited from the first level to the second level that is from n = 1 orbit to n = 2 orbits, it must absorb a photon of ultraviolet light.
But,When an electron falls from n = 2 orbit to n = 1 orbit or from n = 2 orbit(excited state) to n = 0 orbit(groubd state), it emits a photon of ultraviolet light.

7 0

3 years ago

You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standar

djverab [1.8K]

1.04 ⨯ 10 $^{-4}$ M

<h3>Explanation</h3>

<em>A</em> = <em>ε</em> $\cdot l \cdot c$ by the Beer-Lambert law, where

<em>A</em> the absorbance,
$l$ the path length,
<em>ε</em> the molar absorptivity of the solute, and
$c$ concentration of the solution.

<em>A</em> and <em>ε </em>are the same for both solutions. Therefore, $l \cdot c$ is constant; $l$ is inversely proportional to $c$ . The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.

The 13.0 mL standard solution has a concentration of 5.17 ⨯ $10^{-4}$ M. Diluting it to 45.0 mL results in a concentration of $5.17 \times 10^{-4} \times \frac{13.0}{45.0} =$ 1.494 M.

$c$ is inversely related to $l$ for the two solutions. As a result, c₂ = $c_1 \cdot \frac{l_1}{l_2} = 1.494 \times 10^{-4} \times \frac{1}{4.78} =$ 3.126 M.

The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of $3.126 \times \frac{100.0}{30.0} =$ 1.04 ⨯ $10^{-4}$ M.

6 0

3 years ago

If the moles and volume of a gas are held constant, dropping the temperature from 40 degrees Celsius to 20 degrees Celsius cause

evablogger [386]

It would cause a drop <span>but I am not sure double check other answers </span>

8 0

3 years ago

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