in some applications nickel-cadmium batteries have been replaced by nickel-zinc batteries a single nickel-cadmium cell has a vol
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Answer:
- <u>Option b. Atom P has an estimated Zeff of 7 and is therefore to the right of Atom Q, which has a Zeff of 6.</u>
Explanation:
Please, find attached the figures of both atom Q and atom P corresponding to this question.
The <u>features of atom Q are</u>:
- Each <em>black sphere</em> represents an electron
- In total this atom has 8 electrons: 2 in the inner shell and 6 in the outermost shell.
- Since it is assumed that the atom is neutral, it has 8 protons: one positive charge of a proton balances one negative charge of an electron. Thus, the atomic number of this atom is 8.
- Since only two shells are ocuppied, you can assert that the atom belongs to the period 2 (which is confirmed looking into a periodic table with the atomic number 8).
- <em>Zeff </em>is the effective nuclear charge of the atom. It accounts for the net positive charge the valence electrons experience. And may, in a very roughly way, be estimated as the number of protons less the number of electrons in the inner shells. Thus, for this atom, an estimated Z eff = 8 - 2 = 6.
The <u>features of atom P</u> are:
- Again, each black sphere represents an electron
- In total this atom has 9 electrons: 2 in the inner shell and 7 in the outermost shell.
- Since it is assumed that the atom is neutral, it has 9 protons.
- The atomic number of this atom is 9.
- Using the same reasoning used for atom Q, this atom is also in the period 2.
- Estimated Z eff = 9 - 2 = 7.
Then, since atom P has a greater Z eff than atom Q (an estimated Zeff of 7 for atom P against an estimated Z eff of 6 for atom Q), and both atoms are in the same period, you can affirm that <em>atom P</em> has a greater atomic number and<em> is therefore to the right of atom Q</em>.
Answer:
1,812 wt%
Explanation:
The reactions for this titration are:
2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻
I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻
The moles in the end point of S₂O₃⁻ are:
0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:
4,463x10⁻⁴ moles of S₂O₃⁻× = 2,2315x10⁻⁴ moles of I₃⁻
As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:
2,2315x10⁻⁴ moles of I₃⁻× = 4,463x10⁻⁴ moles of Ce(IV). These moles are:
4,463x10⁻⁴ moles of Ce(IV)× = <em>0,0625 g of Ce(IV)</em>
As the sample has a 3,452g, the weight percent is:
0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>
I hope it helps!