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zmey [24]
3 years ago
8

Deduce the structure of an unknown compound using the data below: Molecular Formula: C6H12O IR: 1705 cm-1 1H NMR: no absorptions

greater than δ 3 ppm 13C NMR: δ 24.4, δ 26.4, δ 44.2, and δ 212.6 ppm. Resonances at δ 44.2 and 212.6 have very low intensity.

Chemistry
1 answer:
Novosadov [1.4K]3 years ago
7 0

Answer:

The compound elucidated from the spectral data is <u>4-methyl penta-2-none</u>

Explanation:

  • <u>1700 cm-1 from IR data</u> suspects aldehyde/ketone or carboxylic acid. However,since the peak is not a stretched vibration, it implies an aldehyde or ketone present.
  • <u>3 ppm H NMR</u> confirms O-CH3 bond
  • <u>24.4 13C NMR</u> confirms CH3-R bond
  • 26.4 13C NMR  confirms H3C-(R)C(H)-R
  • 44.2 13C NMR Confirms C=
  • 2126 13C NMR confirms aldehyde C=O bond

<u>The deduced structure is 4-methyl penta-2-one (see attached) </u>given multiple CH3 atoms.

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3 years ago
A solution of rubbing alcohol is 68.6 % (v/v) isopropanol in water. How many milliliters of isopropanol are in a 88.2 mL sample
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From the information given, the total volume of  rubbing alcohol is 88.2 ml

68.6 % of this volume is isopropanol. 

We will assume 88.2 ml represents 100% volume, so the volume of water will be 31.4 %

The volume of isopropanol is

68.6/100 x 88.2 → 0.686 × 88.2 = 60.505 ml

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Volume of water will be 88.20 - 60.5 = 27.7 ml

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Adding 60.5 ml of isopropanol to 27.7 ml of water to make up 88.2 ml will give 68.6 % v/v isopropanol to water  solution.



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3 years ago
Explain how and the type of a bond that forms between sodium and chlorine in sodium chloride (NaCl)
Alexandra [31]

Explanation:

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<em>Now ionic bonds areIons are formed by atoms that have non-full outermost electron shells in order to become more like the noble gases in Group 8 of the Periodic Table,</em>

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7 0
2 years ago
The rate of disappearance of HBr in the gas phase reaction2HBr(g) → H2(g) + Br2(g)is 0.301 M s-1 at 150°C. The rate of appearanc
jok3333 [9.3K]

Answer: 0.151

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2HBr(g)\rightarrow H_2(g)+Br_2(g

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}

Given: -\frac{d[HBr]}{dt}]=0.301

Putting in the values we get:

\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}

+\frac{1d[Br_2]}{dt}=0.151

Thus the rate of appearance of Br_2 is 0.151

8 0
3 years ago
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