Answer:- 27.7 grams of
are produced.
Solution:- The balanced equation is:

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.
Molar mass of
= 207.2+2(126.90) = 461 gram per mol
let's do the calculations for the grams of the product for the given grams of each of the reactant:

= 

= 
From above calculations, NaI gives least amount of
, so the answer is, 27.7 g of
are produced.
Answer:
D is the correct answer.
Explanation:
I took the test on A pex. Everyone make sure you look at your answers. They are NOT always in the same order.
Answer: Option C. p-dichlorobenzene and 1,4-dichlorobenzene.
Explanation:
A line-angle formula with six vertices and a circle inscribed corresponds to the compound known as benzene.
Further, according to the IUPAC standards for naming benzene derivatives, you must first number the position of the substituent. In this case, the substituents (chloros) are located at the positions 1 and 4; also, for the benzene derivatives when they have 2 substituents and the positions are 1 and 4, this configuration is known as <em>para </em>or <em>p </em>configuration.
Additionally, this compound has 2 substituents (chloros) so you have to indicate this number (di).
Therefore, the correct answer is C. p-dichlorobenzene and 1,4-dichlorobenzene.
Answer:
the difference is tyat eruptions of less gassy and more gassy is that the less gassy doesnt retain as much gas as the more gassy one and thus the eruption of the less gassy is less damage to the more gassy
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.