Calculate the mass of water vapour present in a room of volume 400 m3 that contains air at 27 °C on a day when the relative humi
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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.
Answer:
Compound B has greater molar mass.
Explanation:
The depression in freezing point is given by ;
..[1]
Where:
i = van't Hoff factor
= Molal depression constant
m = molality of the solution
According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.
The depression in freezing point of solution with A solute:
Molar mass of A =
The depression in freezing point of solution with B solute:
Molar mass of B =
As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.
This means compound B has greater molar mass than compound A,