Answer:
b. have the same kind number of complete shells
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).
A neutralization process is a reaction between an acid and a base which yields a salt and water. From the choices, a neutralization reaction would be: 2 HBr + Ca(OH)2 yields CaBr2 + 2 H2O. Moreover, <span>NH3 + HCl yields NH4Cl is also a neutralization reaction. The complete reaction is actually NH4OH + HCl --> NH4Cl + H2O. NH4OH is the aqueous solution of NH3. This reaction is still a neutralization reaction.
On the other hand, the reaction </span><span>HCl + HBr yields H2 + ClBr is not valid. There is no reaction between HCl and HBr because both are strong acids. They would just dissociate into ions like H+, Cl- and Br-.
The valid reaction that is clearly not a neutralization process is </span><span>H2 + Br2 yields 2 HBr. This is a combination reaction yielding a strong acid HBr.</span>
The reason for this is because nonmetals, have close to fulfilling an octet and need to gain few more electrons to do this, not to lose more. Nonmetals, because of the fact they need only few more electrons to satisfy their octet they would receive or share electrons to do this.
The property that nonmetals have are that they are very electronegative, they possess a strong affinity to pull electron density closer, because they possess fewer electron shells and possess even protons this allows for this.
Answer:
Please, see attached two figures:
- The first figure shows the solutility curves for several soluts in water, which is needed to answer the question.
- The second figure shows the reading of the solutiblity of NH₄Cl at a temperature of 60°C.
Explanation:
The red arrow on the second attachement shows how you must go vertically from the temperature of 60ºC on the horizontal axis, up to intersecting curve for the <em>solubility</em> of <em>NH₄Cl.</em>
From there, you must move horizontally to the left (green arrow) to reach the vertical axis and read the solubility: the reading is about in the middle of the marks for 50 and 60 grams of solute per 100 grams of water: that is 55 grams of grams of solute per 100 grams of water.
Assuming density 1.0 g/mol for water, 10 mL of water is:
Thus, the solutibily is:
