Answer:
THE NEW VOLUME AT 1 K IS 1.2 L
Explanation:
Using Charles' law which states that the volume of a given gas is directly proportional to its temperature provided the pressure remains constant.
Mathematically written as;
V1/T1 = V2/T2 at constant pressure
V1 = Initial volume = 6L
T1 = initial temperature = 5K
T2 = Final temperature = 1K
V2 = final volume = unknown
Re-arranging the equation by making V2 the subject of the equation, we obtain;
V2 = V1 T2 / T1
V2 = 6 * 1 / 5
V2 = 1.2 L
The new volume of the gas sample at 1 K is 1,2 L
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:
Molar solubility of AgBr = 51.33 × 10⁻¹³
Explanation:
Given:
Amount of NaBr = 0.150 M
Ksp (AgBr) = 7.7 × 10⁻¹³
Find:
Molar solubility of AgBr
Computation:
Molar solubility of AgBr = Ksp (AgBr) / Amount of NaBr
Molar solubility of AgBr = 7.7 × 10⁻¹³ / 0.150
Molar solubility of AgBr = 51.33 × 10⁻¹³
Sulfur - S-16
Electronic configuration:
1s2 2s2 2p6 3s2 3p4
Hope it helped!!
