Answer
is: The molar solubility of calcium phosphate is 108s⁵ = Ksp.
<span>
Balanced chemical reaction: Ca</span>₃(PO₄)₂(s) → 3Ca²⁺(aq) + 2PO₄³⁻(aq).<span>
[Ca²</span>⁺] =
3s(Ca₃(PO₄)₂) =
3s.<span>
[PO</span>₄³⁻] = 2s.<span>
Ksp = [Ca²</span>⁺]³ · [PO₄³⁻]².<span>
Ksp = (3s)³ · (2s)².
Ksp = 108s</span>⁵.
s = ⁵√(Ksp ÷ 108).
Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:

The half-cell reactions are:
Oxidation half reaction (anode): 
Reduction half reaction (cathode): 
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BPb%5E%7B2%2B%7D%5D%7D)
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = 
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = +0.63 V
= cell potential for the reaction = ?
= 3.5 M
= 
Now put all the given values in the above equation, we get:


Therefore, the cell potential for this reaction is 0.50 V
Answer:
32.6%
Explanation:
Equation of reaction
2KClO₃ (s) → 2KCl (s) + 3O₂ (g)
Molar mass of 2KClO₃ = 245.2 g/mol ( 122.6 × 2)
Molar volume of Oxygen at s.t.p = 22.4L / mol
since the gas was collected over water,
total pressure = pressure of water vapor + pressure of oxygen gas
0.976 = 0.04184211 atm + pressure of oxygen gas at 30°C
pressure of oxygen = 0.976 - 0.04184211 = 0.9341579 atm = P1
P2 = 1 atm, V1 = 789ml, V2 = unknown, T1 = 303K, T2 = 273k at s.t.p
Using ideal gas equation
=
V2 =
V2 = 664.1052 ml
245.2 yielded 67.2 molar volume of oxygen
0.66411 will yield =
= 2.4232 g
percentage of potassium chlorate in the original mixture =
= 32.6%
The diameters of the spheres are 72.5 meters square
Answer:
0.000000540
Explanation:
Step 1: Make an ICE chart for the solution of AgBr
"S" represents the molar solubility of AgBr
AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
Step 2: Write the expression for the solubility product constant (Ksp)
Ksp = [Ag⁺] [Br⁻] = S × S
Ksp = S² = (0.0007350)² = 0.000000540