Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
Answer:
Color change
Explanation:
If it is a chemical change, then the color will change. Sometimes chemical changes also cause fizzing and bubbling.
Salt lowers the freezing/melting point of water, so in both cases the idea is to take advantage of the lower melting point. Ice forms when the temperature of water reaches 32 degrees Fahrenheit (0 degrees Celsius).
Answer:
0.35 milli moles of ethanol can be theoretically be produced under these conditions.
Explanation:
Moles of glucose = 
milli mole
Moles of ADP = 0.35 milli mole
Moles of Pi = 0.35 milli mole
Moles of ATP = 0.70 milli mole
As we can see that ADP and Pi are in limiting amount which means tat they are limiting reagent. So, the moles of ethanol produced will depend upon the moles of ADP and Pi.
According to reaction, 2 moles of ADP gives 2 moles of glucose.
Then 0.35 milli moles of ADp will give :
of ethanol
0.35 milli moles of ethanol can be theoretically be produced under these conditions.
