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.
Explanation:
a) HNO2(aq) = HNO3(aq) + H2O(l) +NO(g)
b) SoCl2 (l) + H2O (l) = So2(g) + 2HCl(aq)
c) CH4 (g) + 2O2(g) = Co2 (g) + 2H2O(g)
d) 3CuO(s) + 2NH3 (g) = 3Cu(s) + 3H2O (l) + N2(g)
The answer is d. particles move from an area of high concentration to low concentration.
Question:
Sulfuric acid was once produced through the reaction of sulfur trioxide with water. Sulfur trioxide can form through the reaction of sulfur dioxide and oxygen gas. When nitrogen monoxide gas is added to the system, the reaction speeds up significantly because it proceeds through the following steps:
equations
Identify the catalyst in this reaction, explain how you know it is the catalyst, and describe how it increases the rate of the reaction.
Answer:
NO
It is present but not consumed
NO Lowers the activation energy of the reaction
Explanation:
A catalyst is a substance that is present in a chemical reaction and enables the reaction to occur at a faster rte but does not take part n the reaction
Therefore, whereby NO is not consumed, it is the catalyst
It functions by lowering the activation energy
Calculate the ratio by using Henderson-Hasselbalch equation:
pH = pKa + log [neutral form] / Protonated form
3.05 = 2.21 + log [neutral form] / [Protonated form]
3.05 - 2.21 = log [neutral form] / [Protonated form]
0.84 = log [neutral form] / [Protonated form]
[neutral form] / [protonated form] = anti log 0.84 = 6.91