The ________ orbital is degenerate with 5py in a many-electron atom.
<h2>
5px is the correct answer</h2>
The balanced equation for the above reaction is
2Al + 3CuCl₂ --> 2AlCl₃ + 3Cu
stoichiometry of Al to CuCl₂ is 2:3
limiting reactant is when the reactant is fully consumed in the reaction therefore amount of product formed depends on amount of limiting reactant present.
number of Al moles - 0.5 g / 27 g/mol = 0.019 mol
number of CuCl₂ moles - 3.5 g / 134.5 g/mol = 0.026 mol
if Al is the limiting reactant
if 2 mol of Al reacts with 3 mol of CuCl₂
then 0.019 mol of Al reacts with - 3/2 x 0.019 = 0.029 mol of CuCl₂
but only 0.026 mol of CuCl₂ is present
therefore CuCl₂ is the limiting reactant
and 0.026 mol of CuCl₂ reacts with - 0.026/3 x 2 = 0.017 mol of Al is required
but 0.019 mol of Al is present
therefore CuCl₂ is the limiting reactant and Al is in excess
The specific heat : c = 0.306 J/g K
<h3>Further explanation</h3>
Given
Heat = 35.2 J
Mass = 16 g
Temperature difference : 7.2 K =
Required
The specific heat
Solution
Heat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
Input the value :
c = Q / m.∆T
c = 35.2 / 16 x 7.2
c = 0.306 J/g K
Answer:
HOAc is stronger acid than HClO
ClO⁻ is stronger conjugate base than OAc⁻
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
Explanation:
Assume 0.10M HOAc => H⁺ + OAc⁻ with Ka = 1.8 x 10⁻⁵
=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺
Assume 0.10M HClO => H⁺ + ClO⁻ with Ka = 3 x 10⁻⁸
=> [H⁺] = √(3 x 10⁻⁸)(0.10)M = 5.47 x 10⁻⁵M H⁺
HOAc delivers more H⁺ than HClO and is more acidic.
Kb = Kw/Ka, Kw = 1 x 10⁻¹⁴
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
