Two Half Cell reactions are as,
Reduction: PbO₂ + 2e⁻ → Pb²⁺
As it is acidic medium reaction, so add H⁺ on reactant side as it has more number of oxygen atoms and add H₂O on product side, and balance tehe elements, So,
4 H⁺ + PbO₂ + 2e⁻ → Pb²⁺ + 2 H₂O (1)
Oxidation: I⁻ → I₂ + 1e⁻
Balance Number of elements along with electrons,
2 I⁻ → I₂ + 2e⁻ (2)
Compare eq. 1 and 2,
4 H⁺ + PbO₂ + 2e⁻ → Pb²⁺ + 2 H₂O
2 I⁻ → I₂ + 2e⁻
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻
PbO₂ + 2 I⁻ + 4 H⁺ → Pb²⁺ + I₂ + 2H₂O
Number of Coefficients = 11
Answer:
Generally, the atomic radius decreases across a period from left to right and increases down a given group. ... This causes the atomic radius to decrease. Moving down a group in the periodic table, the number of electrons and filled electron shells increases, but the number of valence electrons remains the same.
An energy level of a certain atom is further divided into the orbitals. There are four (4) orbitals. These are the s orbitals, p orbitals, d orbitals, and f orbitals. They are already arranged above as to how they are to be filled. The number of electrons that each can hold are as follows:
s orbitals = 2 electrons
p orbitals = 6 electrons
d orbitals = 10 electrons
f orbitals = 14 electrons
Answer:
8 drops of 1.00 M NaOH will be needed.
Explanation:
Concentration of ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
in bleach solution = 0.02 M
![NaOH=[OH^-]=0.02M](https://tex.z-dn.net/?f=NaOH%3D%5BOH%5E-%5D%3D0.02M)
Concentration of bleach solution we want ,
= 0.02 M
Volume of the bleach solution,
= 20 ml
Concentration of NaOH solution,
= 1.00 M
Volume of the NaOH solution required ,
= ?
1 mL = 20 drops
0.4 mL = 0.4 × 20 drops = 8 drops
8 drops of 1.00 M NaOH will be needed.
