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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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how many grams of solid silver nitrate would you need to prepare 200.0 mL of a 0.150 M AgNO3 solution
aleksklad [387]
0.150 M AgNO3 = x mol / 0.200 Liters
x mol = 0.03 mol AgNO3
0.03 mol AgNO3 (169.9g AgNO3 / 1 mol AgNO3) We are converting moles to grams here with stoichiometry.

Final answer = 5.097 grams, but if you want it in terms of sig figs then it is 5.09 grams.
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3 years ago
Match the alkane names and structures.<br> -butane<br> -methane<br> -ethane<br> -propane
SpyIntel [72]

Answer:

Explanation:

Here is the names and matches of butane methane ethane propane

7 0
3 years ago
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What is true when an element is oxidized? It bonds with the hydroxide ion. It loses electrons to another element. It reacts with
Mama L [17]

Answer: It loses electrons to another element.

Explanation:- Oxidation is the process in which an element loses electrons and there is an increase in the oxidation state. On losing electrons it combines with a electronegative element such as oxygen, sulphur or nitrogen etc.

Fe\rightarrow Fe^{2+}+2e^-

Reduction is the process in which an element gains electrons and there is a decrease in the oxidation state.

\frac{1}{2}O_2+2e^-\rightarrow O^{2-}

8 0
3 years ago
Explain why photosynthesis and cellular respiration are the “real circle of life”
BabaBlast [244]

Explanation:Photosynthesis is important because the earth and all of the creatures in the sea and land need the plants to be able to feed themselves with the help of the sun and with cellular respiration, those plants are able to take in the carbon dioxide that we give and make into oxygen with we need. there fore without them we would not be able to exist.

4 0
3 years ago
Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0
2 years ago
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