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2 years ago

Functional group and bond hybridization of vanillin

Chemistry

1 answer:

lana66690 [7]2 years ago

7 0

Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.

See attached figure for the structure.

Vanillin have 3 functional groups:

1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen

2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring

3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons

Now for the hybridization we have:

The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a sp² hybridization because they are involved in double bonds.

The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a sp³ hybridization because they are involved only in single bonds.

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A rod, X has a positive charge of 8. An otherwise identical rod, Y has a negative charge of 4. The rods are touched together, an

gtnhenbr [62]

Answer:

1. electrons

2. From "Y" to "X"

Explanation:

1. Electrons move between the rod since the electrons are the only charge carriers which are free to move.

2. The particles move from from "Y" to "X" since the electrons are the only charge carriers which are free to move. The positive charge on rod x is due to a deficit of electrons while the negative charge on rod Y is due to the excess of electrons. When the rods come together, the electrons move from "Y" to "X" since the electrons are the only charge carriers which are free to move.

4 0

1 year ago

What is the molarity of a naoh solution if 11.9 ml of a 0.220 m h2so4 solution is required to neutralize a 25.0-ml sample of the

bulgar [2K]

0.355M x 0.0282L= 0.01 moles of H2SO4. Remember sulphuric acid is diprotic so it will release 2 from each molecule.
So moles of protons = 0.01 x 2 = 0.02 moles of H+ 
For neutralization: moles H+ = moles OH- 
Therefore moles of NaOH = 0.02 
conc = moles / volume 
Conc NaOH = 0.02 / 0.025L = 0.8M

4 0

2 years ago

Read 2 more answers

Write the Lewis structure for ethanol (CH3CH2OH), the alcohol found in alcoholic beverages, then answer the following questions:

hodyreva [135]

Answer :

(1) The number of valence electrons present in the compound is, 20

(2) The number of bonded electrons present in the compound is, 16

(3) The number of lone pair electrons present in the compound is, 4

(4) The number of single bonds present in the compound is, 8

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $CH_3CH_2OH$

As we know that carbon has '4' valence electrons, hydrogen has '1' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in $CH_3CH_2OH$ = 2(4) + 6(1) + 6 = 20

According to Lewis-dot structure, there are 16 number of bonding electrons and 4 number of non-bonding electrons or lone pair of electrons.

The Lewis-dot structure of $CH_3CH_2OH$ is shown below.

8 0

2 years ago

Be sure to answer all parts. For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g)

Liono4ka [1.6K]

Answer:

a. -268.13 J/K

b. -279.95 J/K

c. + 972.59 J/K

Explanation:

The value of the change in entropy (ΔS°) can be calculated by:

ΔS° = ∑n*S° products - ∑n*S° reactants, where n is the stoichiometric number of moles.

The values of S° for each substance can be found on a thermodynamic table.

a. 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)

S°, NO2(g) = 240.06 J/mol.K

S°, H2O(l) = 69.91 J/mol.K

S°, HNO3(l) = 155.60 J/mol.K

S°, NO(g) = 210.76 J/mol.K

ΔS° = (210.76 + 2*155.60) - (3*240.06 + 69.91)

ΔS° = -268.13 J/K

b. N2(g) + 3F2(g) → 2NF3(g)

S°, N2(g) = 191.61 J/mol.K

S°, F2(g) = 202.78 J/mol.K

S°, NF3(g) = 260.0 J/mol.K

ΔS° = (2*260.0 ) - (191.61 + 3*202.78)

ΔS° = -279.95 J/K

c. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)

S°, C6H12O6(s) = 212 J/mol.K

S°, O2(g) = 205.138 J/mol.K

S°, CO2(g) = 213.74 J/mol.K

S°, H2O(g) = 188.83 J/mol.K

ΔS° = (6*213.74 + 6*188.83) - (212 + 6*205.138)

ΔS° = +972.59 J/K

3 0

2 years ago

What is the final pressure (expressed in atm) of a 3.05 l system initially at 724 mm hg and 298 k, that is compressed to a final

krok68 [10]

Answer:

P2 = 778.05 mm Hg = 1.02 atm

Explanation:

We are to find the final pressure (expressed in atm) of a 3.05 liter system initially at 724 mm hg and 298 K which is compressed to a final volume of 2.60 liter at 273 K.

For this, we would use the equation:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

where P1 = 724 mm hg

V1 = 3.05 L

T1 = 298 K

P2 = ?

V2 = 2.6 L

T2 = 173 K

Substituting the given values in the equation to get:

$\frac{(724)(3.05)}{298} =\frac{P_2(2.6)}{173}$

P2 = 778.05 mm Hg = 1.02 atm

7 0

2 years ago