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.
<span><span>The reaction is as follows:
C6H6 </span>+ HNO3 + H2SO4 ------------> </span>C6H5NO2<span> + H</span>2<span>O
(BENZENE) (NITRIC ACID)(CATALYST)
</span>NO2(+) is the electrophile that acctacks on the benzene ring in nitration process.
Answer:
The answer to your question is M = 36.49 g
Explanation:
Data
mass = 8.21 g
volume = 4.8064 L
Temperature = 200°C
Pressure = 1.816 atm
M = ?
Process
1.- Convert temperature to °K
°K = 273 + 200
°K = 473
2.- Calculate the number of moles
n = (PV)/RT
n = (1.816)(4.8064)/(0.082)(473)
n = 0.225
3.- Calculate the molar mass
M --------------- 1 mol
8.21 g ---------- 0.225 moles
M = (1 x 8.21)/0.225
M = 36.49 g
Answer:
13.94moles of Na₂O
Explanation:
The balanced reaction expression is given as:
4Na + O₂ → 2Na₂O
Given parameters:
Number of moles of O₂ = 6.97moles
Unknown:
Number of moles of Na₂O
Solution:
To solve this problem;
1 mole of O₂ will produce 2 moles of Na₂O ;
6.97 moles of O₂ will produce 6.97 x 2 = 13.94moles of Na₂O
Answer:
313, 6grams of H3PO4
Explanation:
We calculate the weight of 1 mol of H3PO4:
Weight 1 mol H3PO4= (Weight H)x3+ (Weight P)+(Weight 0)x4 =1gx3+31g+16gx4
Weight 1 mol H3PO4=98 g /mol
1 mol-----98 grams H3PO4
3,2mol----x= (3,2molx 98 grams H3PO4)/ 1mol=313,6 grams H3PO4
