Functional group and bond hybridization of vanillin
1 answer:
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 <u>sp²</u> 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 <u>sp³</u> hybridization because they are involved only in single bonds.
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Answer:
The temperature of the gas is 876.69 Kelvin
Explanation:
Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P*V = n*R*T
where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.
In this case:
- P= 470 mmHg
- V= 570 mL= 0.570 L
- n= 0.216 g= 0.0049 moles (being the molar mass of carbon dioxide is 44 g/mole)
- R= 62.36367
- T=?
Replacing:
470 mmHg*0.570 L= 0.0049 moles* 62.36367 *T
Solving:
T= 876.69 K
<em><u>The temperature of the gas is 876.69 Kelvin</u></em>