Answer:
a. electrophilic aromatic substitution
b. nucleophilic aromatic substitution
c. nucleophilic aromatic substitution
d. electrophilic aromatic substitution
e. nucleophilic aromatic substitution
f. electrophilic aromatic substitution
Explanation:
Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).
A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).
You multiply avogadro's number to what you were given.
8.30x10^23 * 6. 0221409x10^23
=1.357*10^25
That should be the right answer but I'm not sure. It has been awhile since I have done this.
Answer:
Empirical formula is Li₂CO₃.
Explanation:
Percentage of oxygen= 65.0%
Percentage of lithium = 18.7%
Percentage of carbon= 16.3%
Empirical formula = ?
Solution:
Number of gram atoms of C = 16.3/12 = 1.4
Number of gram atoms of Li = 18.7/6.94 = 2.7
Number of gram atoms of O = 65.0/ 16 = 4.1
Atomic ratio:
Li : C : O
2.7/1.4 : 1.4/1.4 : 4.1/1.4
2 : 1 : 3
Li : C : O = 2 : 1 : 3
Empirical formula is Li₂CO₃.
Answer:
20% of phosphorus
Explanation:
A fertilizer is used to improve the fertility of soils. Most fertilizers contains the element nitrogen, phosphorus and potassium.
They are often designated NPK fertilizers.
Now we know that the numbers 10-20-20 depicts the nitrogen-phosphorus and potassium content of the fertilizer.
From the designation,
The actual percentage is 20% of phosphorus.
10% of nitrogen
20% of potassium
Answer:
B
Explanation:
Diamagnetism , paramagnetism and ferromagnetism are terms which are used to describe the magnetic activity of transition metals. These terms helps to know the response the metal will have when placed in a magnetic field. These activities can be discerned from the d-block electronic configuration. If the number of electrons are even, this means they all form a pair and the metal is diamagnetic. If otherwise, there is an unpaired electron which causes the paramagnetic activity of the metal in magnetic field.
To the question, options A-D are diamagnetic but configurations with d8 are the ones that form square planar complexes
