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irina1246 [14]
3 years ago
14

33. Hydrocarbons that release pleasant odors are called_________ hydrocarbons. (1 point)

Chemistry
2 answers:
olya-2409 [2.1K]3 years ago
8 0

Answer:

Aromatic Hydrocarbons

Explanation:

Aromatic (Pleasant Odour) Hydrocarbons are those having pleasant odours.

tresset_1 [31]3 years ago
6 0

Answer:

substituted hydrocarbons

Explanation:

i think

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Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:
Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

                                   \frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} ),

where  λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from n_1 to n_2 and R_h = 109677 is the Rydberg Constant. Here n_1 and  n_2 represents the transitions.

(a) n_1 =2 to n_2 = infinity

            \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2}  ) = 109677/4     [since 1/infinity = 0] Therefore, \lambda = 4 / 109677 = 0.00003647 m

(b) n_1=4 to  n_2 = 20

           \frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2}  ) = 6580.62

Therefore,  \lambda = 1 / 6580.62 = 0.000152 m

(c) n_1=3 to  n_2 = 10

          \frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2}  ) = 11089.56

Therefore,  \lambda = 1 / 11089.56 = 0.00009 m

(d)  n_1=2 to  n_2 = 1

          \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2}  ) = - 82257.75

Therefore,  \lambda = 1 /82257.75  = - 0.0000121 m  

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

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8 0
1 year ago
List three properties of metal that nonmetals typically DO NOT have
finlep [7]
-They can conduct heat
-They can conduct electricity
-They are typically stronger than non metals
Hope this helps, have a nice day! :)
7 0
3 years ago
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Calculate the molarity of a solution prepared by dissolving 11.9 g of beryllium chloride in a total volume of 292.2 mL of soluti
anyanavicka [17]

Answer:

8793

Explanation:

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3 years ago
If you mix cu2+ with? a. nacl, b. nano2, c. h2o , arrange the solutions based on their absorption from highest frequency to lowe
erica [24]

The arrangement of the solutions based on their absorption from highest frequency to lowest frequency :

b.NaNO_{2} > c.H_{2} O > a.NaCl

<h3>What is absorption frequency?</h3>
  • The frequency of the molecular vibration that led to the absorption is the same as the absorption frequency of a basic IR absorption band.
  • In a way, an emission spectrum is the opposite of an absorption spectrum.
  • The discrepancies in the energy levels of each chemical element's orbitals correspond to absorption lines for each chemical element at various particular wavelengths.
  • Therefore, it is possible to identify the constituents in a gas or liquid using its absorption spectrum.
  • Absorption spectroscopy is most frequently used to measure infrared, atomic, visible, ultraviolet (UV), and x-ray waves.

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3 0
1 year ago
If one-trillionth of the atoms of a radioactive isotope disintegrate each day, what is the decay constant of the process?
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Decay constant of the process 1×10^(-12) day^(-1).

<h3>What is decay constant?</h3>

A radioactive nuclide's probability of decay per unit time is known as its decay constant, which is expressed in units of s1 or a1. As a result, as shown by the equation dP/P dt =, the number of parent nuclides P declines with time t. Nuclear forces are about 1,000,000 times more powerful than electrical and molecular forces in their ability to bind protons and neutrons. The strength of the bonds holding the radioactive element are likewise indifferent to the decay probabilities and's, in addition to being unaffected by temperature and pressure. The decay constant is related to the nuclide's T 1/2 half-life by T 1/2 = ln 2/.

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2 years ago
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