Answer:
0.12 mol KCl
Explanation:
2 KClO3 (s) 2 KCl (s) + 3 O2 (g)
15 g x mol
x g KCl = 15 g KClO3 x[ (1 mol KClO3)/ (122.5 g KClO3) ] x [(2 mol KCl)/ (2 mol KClO3)]
x g KCl = 0.12 mol KCl
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).
Ksp of AgCl= 1.6×10⁻¹⁰
AgCl=Ag⁺ +Cl⁻
Ksp=[Ag⁺][Cl⁻]
Assume [Ag⁺]=[Cl⁻]=x
Ksp=x²
1.6×10⁻¹⁰=x²
x=0.000012
In FeCl₃:
FeCl₃------>Fe⁺³+ 3Cl⁻
as there is 0.010 M FeCl₃
So there will be ,
[Cl⁻]= 0.030
So
[Ag⁺]=Ksp/[Cl⁻]
=1.6×10⁻¹⁰/0.030
=5.3×10⁻⁹
so solubility of AgCl in FeCl₃ will be 5.3×10⁻⁹.
Answer:Yes they are in the same mineral group
Explanation:zinc is the central elements there. The rest of the elements are present as impurities due to where it was found. Like carbon is can be found in the soil, silicon with oxygen is basically sand, hydrogen is in the atmosphere and also in water and soil too. So apart from zinc, the rest are normal day to day elements.
Answer:
Pressure = 0.64 atm
Explanation:
Use Ideal Gas Law PV = nRT => P = nRT/V
n = moles CO₂(g) = 66g/44g/mol = 1.5 mol CO₂
R = Gas Constant = 0.08206 L·atm/mol·K
T = Temperature = -14.5°C = (-14.5 + 273)K = 258.5 K
V = 50.0 Liters
P = (1.5 mol)(0.08206 L·atm/mol·K)(258.5 K)/(50.0 L) = 0.64 atm
