B, Frost Wedging. Have a good rest of your day!
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:
287.30 g of FeCO₃
Solution:
The Balance Chemical Equation is as follow,
FeCl₂ + Na₂CO₃ → FeCO₃ + 2 NaCl
Step 1: Calculate Mass of FeCl₂ as,
Molarity = Moles ÷ Volume
Solving for Moles,
Moles = Molarity × Volume
Putting Values,
Moles = 2 mol.L⁻¹ × 1.24 L
Moles = 2.48 mol
Also,
Moles = Mass ÷ M.Mass
Solving for Mass,
Mass = Moles × M.Mass
Putting Values,
Mass = 2.48 mol × 126.75 g.mol⁻¹
Mass = 314.34 g of FeCl₂
Step 2: Calculate Mass of FeCO₃ formed as,
According to equation,
126.75 g (1 mole) FeCl₂ produces = 115.85 g (1 mole) FeCO₃
So,
314.34 g of FeCl₂ will produce = X g of FeCO₃
Solving for X,
X = (314.34 g × 115.85 g) ÷ 126.75 g
X = 287.30 g of FeCO₃
<h2>
brainlyest pleas</h2>
According to Balance chemical equation,
N₂ + 3 H₂ → 2 NH₃
1 mole of Nitrogen reacts with 3 moles of Hydrogen to produce 2 mole of Ammonia.
It is known that i mole of any gas at standard temperature and pressure occupies 22.4 L of Volume. So, we can also say,
22.4 L (1 × 22.4) of Nitrogen gas (in question it is taken in excess) reacts with 67.2 L (22.4 × 3) of Hydrogen gas to produce 44.8 L (22.4 × 2) of Ammonia.
Result:
44.8 L is the correct answer.
Mass number - # of protons = #neutrons, so I would say the answer is 5