Answer:
Cu(s) in Cu(NO₃)₂(aq)
Explanation:
The standard reduction potential (E°) is the energy necessary to reduce the atom in a redox reaction. When an atom reduces it gains electrons from other than oxides. As higher is E°, easily it will reduce. The substance that reduces is at the cathode of a cell, where the electrons go to, and the other that oxides are at the anode of the cell.
The standard reduction potentials from Al(s) and Cu(s) are, respectively, -1.66V and +0.15V, so the half-cell of Cu(s) in Cu(NO₃)₂(aq) is the cathode.
To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration of the stock
solution, V1 is the volume of the stock solution, M2 is the concentration of
the new solution and V2 is its volume.
65 x V1 = 2 x 200 L
V1 = 6.15 L
Answer:
B-2
Explanation:
In the graph you can so that two of the shapes ae fully black, that means that they are color blind, the half colored ones means they are a carrier but they aren't color blind. So two males in the offspring are color blind.
Given the solubility of strontium arsenate is 0.0480 g/l . we have to convert it into mol/L by dividing it over molar mass (540.7 g/mol)
Molar solubility = 0.0480 / 540.7 = 8.9 x 10⁻⁵ mol/L
Dissociation equation:
Sr₃(AsO₄)₂(s) → 3 Sr²⁺(aq) + 2 AsO₄³⁻(aq)
3 s 2 s
Ksp = [Sr²⁺]³ [AsO₄³⁻]²
= (3s)³ (2s)²
= 108 s⁵
Ksp = 108 (8.9 x 10⁻⁵) = 5.95 x 10⁻¹⁹
