Answer is: at higher temperatures reaction will go to the right (forward), more products (C₂H₄ and H₂) will be produce, because this is endothermic reaction (ΔH<span> is positive, </span>energy is consumed) and according Le Chatelier's principle <span>heat is included as a reactant. </span> .
Answer:
5
Explanation:
Given parameters:
Hydrogen ion concentration = 0.00001M
Unknown:
pH of the solution =?
Solution:
The pH is used to estimate the degree of acidity or alkalinity of a solution. To solve for pH of any solution, we use the expression below;
pH = -log [H⁺]
[H⁺] is the hydrogen ion concentration
pH = -log (1 x 10⁻⁵)
pH = -(-5) = 5
Answer:
The resulting solution contains approximately 666 g of water.
Explanation:
In the initial solution we have:
1g salt : 8g sugar : 200g water
This means that the ratios are:
In the final solution we have:
5g salt: xg sugar: yg water
The new ratios are:
Now we can calculate the amount of sugar in the final solution:
Finally, we calculate the amount of water:
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
