It makes an ecosystem stable
probably condensation and evaporation and freezing.. PROBABLY idk
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).
2 moles of NaOH dissolved in 1 litre of solution is the solution with more concentration.
Answer: Option A
<u>Explanation:</u>
Concentration of solution is the measure of the amount of solute dissolved in the solvent of the solution. So this is measured using the molarity of the solution. Molarity is determined as the number of moles of the solute present in the given amount of solvent.
In this present case, the option A gives the molarity of 2 M as
But the second option, mass of NaOH is given. So we have to determine the molarity. First we have to find the molar mass of NaOH. We know that 1 mole of NaOH will contain 40 g/mole.
1 g of NaOH = 40 g of NaOH
1 g of NaOH = 1/40 moles
So 2 g of NaOH will contain 
which is equal to 0.05 moles of NaOH.
Thus, the molarity of 2 g of NaOH will be
Molarity = 
=0.05 M
Thus, the option A is having higher concentration as the molarity is more for 2 moles of NaOH dissolved in 1 l of solution.
