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navik [9.2K]
3 years ago
10

Which of the following groups MAY have no detectable IR absorption in the range usually associated with an absorption?

Chemistry
1 answer:
Svetlanka [38]3 years ago
5 0

Here we have to get the functional group which may not be detected in IR spectrum.

The ketene and alkyne group may not be detectable by IR absorption.

The infrared (IR) spectrum is an important record which gives sufficient information about the structure of a compound. The absorption of infrared radiation causes the various bands in a molecule to stretch and bend with respect to one another. The most important region for an organic chemist is 4000 cm⁻¹ to 667 cm⁻¹. The region is called the IR region.

The IR spectra is an important tool to recognize the presence of a particular group in an organic group.

The IR region of the given functional group are shown as:

1. Alkyne -C≡C- 2100 cm⁻¹  to 2260 cm⁻¹

2. Ketene -C=C=O- 2100 cm⁻¹ to 2150 cm⁻¹

3. Aldehyde carbonyl -CH=O 1700 cm⁻¹ to 1725 cm⁻¹

4. nitrile -C≡N 2200 cm⁻¹ to 2250 cm⁻¹

5. Ether -C-O- 1000 cm⁻¹ to 1300 cm⁻¹.

Among the function groups the -C≡C- bond is hard to detect by IR spectrum. It is not found in symmetrical alkyne.

The ketene group also not clearly indicative in the IR region.

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lakkis [162]
The formula for phosphorus is P4
3 0
3 years ago
Fe2O3 + 2Al -> Al2O3 + 2Fe
pashok25 [27]
Fe2O3 + 2Al ---> Al2O3 + 2Fe 
Mole ratio Fe2O3 : Al = 1:2 
No. of moles of Fe2O3 = Mass/RMM = 250 / (55.8 * 2 + 16 * 3) = 1.56641604 moles 
No. of moles of Al = 150/27 = 5.555555555 moles. 
Mole ratio 1 : 2. 1.56641604 * 2 = 3.13283208 moles of Al, but you have 5.555555555 moles of Al. So Al is in excess. All of it won't react. 

So take the Fe2O3 and Fe ratio to calculate the mass of iron metal that can be prepared. 
RMM of Fe2O3 / Mass of Fe2O3 = RMM of 2Fe / Mass of Fe 159.6 / 250 = 111.6 / x x = 174.8 g of Fe 
7 0
3 years ago
Read 2 more answers
Determine the number of representative particles in each of the following:
tia_tia [17]

Answer:

(a) 0.294 mol silver = 1.770 * 10^{23}  

(b) 8.98 * 10-3 mol sodium chloride - 5.407 *10^{23}  

(c) 23.3 mol carbon dioxide  = 1.403 * 10^{25}

(d) 0.310 mol nitrogen (N2) = 1.866 * 10^{23}

Explanation:

In one mole there are 6.022 * 10^{23} atoms/molecules

(a) 0.294 mol silver = 6.022 * 10^{23} * 0.294 = 1.770 * 10^{23}  

(b) 8.98 * 10-3 mol sodium chloride - 5.407 *10^{23}  

(c) 23.3 mol carbon dioxide  = 23.3 * 6.022 * 10^{23} = 1.403 * 10^{25}

(d) 0.310 mol nitrogen (N2) = 0.310 * 6.022 * 10^{23} = 1.866 * 10^{23}

3 0
3 years ago
A chemist reacted 57.50 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is show
pantera1 [17]
(On my left) (On my right side)
Na: 1. Na:1
Cl:2. Cl:1

Na:1 x 2 =2
Na:1 x 2=2
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Answer:
2Na + Cl2 -> 2NaCl
5 0
3 years ago
What is the mass of one mole of H
arsen [322]

Answer:

1.008 grams

Explanation:

This numbers are either in grams per mole of atoms or atomic mass unit

4 0
3 years ago
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