According to this equation:
AgCl(s) ↔ Ag+(aq) + Cl-(aq)
so K1 = [Ag+][Cl-]
when [Ag+] = [Cl-] we can assume both = X
and when we have X the solubility = 1.33 x 10^-5 mol / L
by substitution:
∴ K1 = X^2
= (1.33 x 10^-5)^2
= 1.77 x 10^-10
by using vant's Hoff equation:
ln(K2/K1) = (ΔH/R)*(1/T2-1/T1)
when ΔH = 65700 J / mol
R = 8.314
T1 = 25+273 = 298 K
T2 = 47.7 +273 =320.7
by substitution:
∴㏑(K2/1.77 x 10^-10) = (65700/8.314) * ( 1/320.7 - 1/ 298)
by solving for K2
∴K2 = 2.7 x 10^-11
and when K2 = X^2
∴ the solubility X = √(2.7 x 10^-11)
= 5.2 x 10^-6 mol/L
Answer:
copper wire, metal nail, battery
Explanation:
To create an electromagnet, the material to be magnetized (such as a metal nail) is placed inside a solenoid, the ends of the coils are connected to a circuit that includes a battery.
Current is allowed to flow for a few seconds and then cut off. The polarity of the magnet depends on the direction of flow of current in the solenoid.
Answer:
Q = 1.08x10⁻¹⁰
Yes, precipitate is formed.
Explanation:
The reaction of CoF₂ with NaOH is:
CoF₂(aq) + 2 NaOH(aq) ⇄ Co(OH)₂(s) + 2 NaF(aq).
The solubility product of the precipitate produced, Co(OH)₂, is:
Co(OH)₂(s) ⇄ Co²⁺(aq) + 2OH⁻(aq)
And Ksp is:
Ksp = 3x10⁻¹⁶= [Co²⁺][OH⁻]²
Molar concentration of both ions is:
[Co²⁺] = 0.018Lₓ (8.43x10⁻⁴mol / L) / (0.018 + 0.022)L = <em>3.79x10⁻⁴M</em>
[OH⁻] = 0.022Lₓ (9.72x10⁻⁴mol / L) / (0.018 + 0.022)L = <em>5.35x10⁻⁴M</em>
Reaction quotient under these concentrations is:
Q = [3.79x10⁻⁴M] [5.35x10⁻⁴M]²
<em>Q = 1.08x10⁻¹⁰</em>
As Q > Ksp, <em>the equilibrium will shift to the left producing Co(OH)₂(s) </em>the precipitate
Explanation:
a. Vanillin(4-hydroxy-3-methoxybenzaldehyde):
In its structure hydroxl group will be present on para position of the benzaldehyde ring and methoxy group on meta position.
b. Thymol (2-isopropyl-5-methylphenol):
In its structure isopropyl group will be present on ortho position of the phenol ring and methyl group on meta position.
c. Carvacrol (5-isopropyl-2-methylphenol):
In its structure isopropyl group will be present opposite to methyl group which is present ortho position in a phenol ring.
d. Eugenol (4-allyl-2-methoxyphenol):.
In its structure allyl group will be present on para position of the phenol ring and methoxy group on ortho position.
e. Gallic acid (3,4,5-trihydroxybenzoic acid):
In its structure hydroxyl group will be present on both meta positions and on para position of the benzoic acid ring.
f. Salicyl alcohol (o-hydroxybenzyl alcohol):
In its structure, 
group is linkedto benzene ring and in respect to that hydroxyl group is present at ortho position of the ring.
