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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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How many moles of carbon are in 7.87*10^7
elena-14-01-66 [18.8K]
Well one mole of stuff, any stuff, including carbon dioxide, specifies
6.022
×
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23
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Explanation:
And thus we work out the quotient:
7.2
×
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25
⋅
carbon dioxide molecules
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×
10
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⋅
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⋅
m
o
l
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carbon dioxide
.
This is dimensionally consistent, because we get an answer with units
1
m
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−
1
=
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as required.
6 0
3 years ago
Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2NH3(g) 3N2O(g)4N2(g) 3H2O
trasher [3.6K]

Answer: \Delta H^{0} = -879.15 kJ/mol

Explanation: <u>Heats</u> <u>of</u> <u>formation</u> is the amount of heat necessary to create 1 mol of a compound from its molecular constituents. The basic conditions the substance is formed is at standard conditions: 1 atm and 25°C. Each compound has its own heat of formation per mol of compound (kJ/mol), but to an element is assigned a value of zero.

<u>Standard</u> <u>Enthalpy</u> <u>Change</u> is defined as the heat absorbed or released when a reaction takes place. It can be positive or negative, which means reaction is endothermic or exothermic, respectively.

Enthalpy change is calculated as the difference between the sum of heat formation of products and the sum of heat formation of the reactants:

\Delta H^{0}=\Sigma H^{0}_{f}_{(products)}-\Sigma H^{0}_{f}_{(reactants)}

For the reaction

   2NH₃   +      3N₂O   →      4N₂    +     3H₂O

2(-46.2)   +     3(82.05)      4(0)    +    3(-241.8)

\Delta H^{0}=3(-241.8)-[ 2(-46.2)+3(82.05)]

\Delta H^{0}=-725.4-153.75

\Delta H^{0}=-879.15

<u>The standard enthalpy change for the reaction is </u>\Delta H^{0}=-879.15<u> kJ</u>

8 0
2 years ago
17. Is the scientific method suitable for solving problems only in<br> the sciences? Explain.
Neporo4naja [7]

Answer:

Hey mate

Explanation:

Yes these methods are only used in biology, chemistry, physics, geology and physcology.... They can't be used in maths or other subject, as it is said SCIENCE IS SCIENCE!

Hope it helps you,

mark me the brainliest,

follow me

6 0
2 years ago
What causes the difference in bond angles in carbon dioxide and water?
egoroff_w [7]

Answer is: C) the fact that the number of lone pairs of electrons on the central atom is greater in the case of water.

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Oxygen atom has atomic number 8, it means it has eight protons and eight electrons, so atom has neutral charge. Oxygen is a nonmetal.

Electron configuration of oxygen atom: ₈O 1s² 2s² 2p⁴.

Oxygen atom has six valence electrons , two lone pairs and two electrons that form two sigma bonds with hydrogen atoms.

Carbon is a chemical element with symbol C and atomic number 6, which means it has 6 protons and six electrons. Four valence electrons are in 2s and 2p orbitals.  

Electron configuration of carbon atom: ₆C 1s² 2s² 2p².  

In carbon dioxide, carban has sp hybridization with no lone pairs.

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Answer:c

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2 years ago
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