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 is: volume of carbon dioxide is 1,84·10⁸ l.
Chemical reaction: C + O₂ → CO₂.
m(C) = 100 t · 1000 kg/t = 100000 kg
m(C) = 100000 kg · 1000 g/kg = 10⁸ g.
n(C) = m(C) ÷ M(C).
n(C) = 10⁸ g ÷ 12 g/mol.
n(C) = 8,33·10⁶ mol.
From chemical reaction: n(C) . n(CO₂) = 1 : 1.
n(CO₂) = 8,33·10⁶ mol.
m(CO₂) = 8,33·10⁶ mol · 44 g/mol.
m(CO₂) = 3,66·10⁸ = 3,66·10⁵ kg.
V(CO₂) = 3,66·10⁵ kg ÷ 1,98 kg/m³ = 1,84·10⁵ m³.
V(CO₂) = 1,84·10⁵ m³ · 1000 l/m³ = 1,84·10⁸ l.
<span> under extreme heat or pressure
crystallize from magma
precipitate from a solution reaction of hot mixture with water and a dissolved substance</span>
Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. ... When an equal number of atoms of an element is present on both sides of a chemical equation, the equation is balanced.