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3 years ago

What happens after excitation in spectroscopy.

Chemistry

1 answer:

Bess [88]3 years ago

4 0

Answer:

When a molecule is excited to a higher state it often ends up in its lowest excited state S1 and then emits radiation.

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How does the Doppler effect related to the universal expanding?

tensa zangetsu [6.8K]

The Doppler effect doesn't just apply to sound. It works with all types of waves, which includes light. Edwin Hubble used the Doppler effect to determine that the universe is expanding. Hubble found that the light from distant galaxies was shifted toward lower frequencies, to the red end of the spectrum. This is known as a red Doppler shift, or a red-shift. If the galaxies were moving toward Hubble, the light would have been blue-shifted.

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3 years ago

A certain hydrocarbon has a molecular formula of C9H16. Which of the following is not a structural possibility for this hydrocar

snow_tiger [21]

Answer: Option (d) is the correct answer.

Explanation:

It is given that molecular formula is $C_{9}H_{16}$ . Now, we will calculate the degree of unsaturation as follows.

Degree of unsaturation = $C_{n} - \frac{\text{monovalent}}{2} + \frac{\text{trivalent}}{2} + 1$

= $9 - \frac{16}{2} + 1$

= 9 - 8 + 1

= 2

As the degree of unsaturation comes out to be 2. It means that this compound will contain one ring and one double bond.

Yes, this compound could be an alkyne as for alkyne D.B.E = 2.

But this compound cannot be a cycloalkane because for a cycloalkane D.B.E = 1 which is due to the ring only.

Thus, we can conclude that it is a cycloalkane is not a structural possibility for this hydrocarbon.

6 0

3 years ago

The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W

Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹

Explanation:

Assume the rate law is

rate = k[A][B]²

If you are comparing two rates,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}$

You are cutting each concentration in half, so

$\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}$

Then,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}$

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3 years ago

1. How many moles of oxygen are made if 12.0 moles of potassium chlorate react?

Alex787 [66]

Answer:

Explanation:

12*3=36

36/2=18

567890

3 0

3 years ago

What is the mass of 22.4 L of H2 at STP?

Vanyuwa [196]

A. 1.01 is the right answer

Since

The formula is Pv= nRT

P=1 atm

V= 22.4 L

N= x

r= 0.0821

t = 273 k (bc it’s standard temperature)

So (1)(22.4)=(x)(0.0821)(273)

X= 1.001

7 0

3 years ago

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What happens after excitation in spectroscopy.​

What happens after excitation in spectroscopy.