When Pb(OH)₂ dissolves it dissociates as follows;
Pb(OH)₂ ---> Pb²⁺ + 2OH⁻
molar solubility is the number of moles of compound that can be dissolved in 1 L of solution.
if molar solubility of Pb(OH)₂ is x then molar solubility of Pb²⁺ is x and OH⁻ is 2x
the formula for solubility product constant is as follows;
ksp = [Pb²⁺][OH⁻]²
ksp = (x)(2x)²
ksp = 4x³
ksp = 1.43 x 10⁻²⁰
4x³ = 1.43 x 10⁻²⁰
x = 1.53 x 10⁻⁷ M
molar solubility of Pb(OH)₂ is 1.53 x 10⁻⁷ M
molar mass is 241.2 g/mol
solubility of Pb(OH)₂ is 1.53 x 10⁻⁷ M x 241.2 g/mol = 3.69 x 10⁻⁵ g/L
Boyle's law: in Boyle's law, pressure of a gas in inversely proportional to the volume while the temperature is shield constant
Gay-Lussac's law: In Gay-Lussac's law, pressure of a gas in directly proportional to the temperature while the volume is held constant.
Charles's law: In Charles's law, the volume of a gas is directly proportional to the temperature while the pressure is held constant.
Answer:
= 1.5 eq
Explanation:
One definition of an equivalent weight is that it is mass of a substance that gains or loses 1 mole of electrons.
Al3+ has lost 3 e-, so there are 3 equivalent weights in 1 mol Al3+.
1 mol Al3+ =3 eq. wts.
1 mol Al x(27 g / 1 mol)x(1 mol / 3 eq. wts.) = 9.0 g = 1 eq. wts.
13.5 g Al3 + x (1 eq.wt. / 9.0 g) = 1.5 eq
Answer:
The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction
The second run has twice the surface area - yes, 44 sqcm to 22 sqcm
Explanation:
A catalyst is a material which speeds up a reaction without being consumed in the process. A heterogeneous catalyst is one which is of a different phase than the reactants. The effectiveness of a catalyst is dependent on the available surface area. The first step for this question is to determine the total available surface area of catalyst in both processes.
Step 1: Determine radius of large sphere




Step 2: Determine surface area of large sphere



Step 3: Determine radius of small sphere




Step 4: Determine surface area of small sphere



Step 5: Determine total surface area of 8 small spheres



- Surface area of 1 large sphere
- Surface area of 8 small spheres
Options:
- The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
- The second run will be slower - false, the increased surface area of catalyst will increase the rate of reaction
- The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction
- The second run has twice the surface area - yes, 44 sqcm to 22 sqcm
- The second run has eight times the surface area - no, 44 sqcm to 22 sqcm
- The second run has 10 times the surface area - no, 44 sqcm to 22 sqcm