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.
Partial pressure (N2) = mole fraction * total pressure
{ 1 mole of any ideal gas occupy same volume of 1 mole of any other ideal gas under same condition of temperature and pressure so mole fraction in the sample is simply 78.08% = 0.7808 this is because equal volume of each gas has equal moles
partial pressure N2 = 0.7808 * 760 .0
partial pressure = 593.4 mmhg ( 1 torr = 1mmhg )
Answer:
8.96 L
Explanation:
At STP, 1 mole = 22.4 L
0.400 mole * (22.4 L. /1 mole of gas) = 8.96 L
Given:3.40g sample of the steel used to produce 250.0 mLSolution containing Cr2O72−
Assuming all the Cr is contained in the BaCrO4 at the end.
(0.145 g BaCrO4) / (253.3216 g BaCrO4/mol) x (250.0 mL / 10.0 mL) x (1 mol Cr / 1 mol BaCrO4) x (51.99616 g Cr/mol / (3.40 g) = 0.219 = 21.9% Cr
Answer:
12 átomos de oxígeno hay presentes
Explanation:
Basados en la reacción:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
<em>6 moles de agua producen 1 mol de glucosa</em>
<em />
Si reaccionan 12 moleculas de agua, se producirán:
12 moleculas H₂O * (1 mol C₆H₁₂O₆ / 6 mol H₂O) =
2 moléculas de glucosa se producen.
Como cada molécula de glucosa tiene 6 átomos de oxígeno:
2 moléculas C₆H₁₂O₆ * (6 átomos Oxígeno / 1 molécula C₆H₁₂O₆) =
<h3>12 átomos de oxígeno hay presentes</h3>
