Answer:
5. The valence electrons of both fluorine and carbon are found at about the same distance from their respective nuclei but the greater positive charge of the fluorine nucleus attracts its valence electrons more strongly.
Explanation:
Both fluorine and carbon are located in the second period of the periodic table, it means that they have 2 shells, so the valence electrons are found at about the same distance from their respective nuclei.
But fluorine has a higher atomic number, 9, than the carbon, 6. The atomic number represents how many protons there are in the nucleus, then there are more protons (positive charge) at the fluorine nucleus, and because of that, the attraction force between the nucleus and the valence electron is stronger in fluorine.
If the force is stronger, it will be necessary more energy to break the bond, so it will be harder to remove an electron from fluorine than from carbon.
Aluminum sulfide : Al₂S₃
ratio cation : anion = 2 : 3
<h3>Further explanation</h3>
Given
Compound of Aluminium
Required
cations anions ratio
Solution
Salt can be formed from cations and anions which have their respective charges.
In the chemical compound formula these charges are crossed with each other
For aluminum it has a +3 charge
1. Aluminum carbide : Al₄C₃
ratio cation : anion = 4 : 3
2. Aluminum chloride : AlCl₃
ratio cation : anion = 1 : 3
3. Aluminum sulfide : Al₂S₃
ratio cation : anion = 2 : 3
4. Aluminum nitride : AlN
ratio cation : anion = 1 : 1
Pb(NO₃)₂ ⇒limiting reactant
moles PbI₂ = 1.36 x 10⁻³
% yield = 87.72%
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)
Required
- moles of PbI₂
- Limiting reactant
- % yield
Solution
Balanced equation :
Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)
mol Pb(NO₃)₂ :
= 0.45 : 331 g/mol
= 1.36 x 10⁻³
mol NaI :
= 250 ml x 0.25 M
= 0.0625
Limiting reactant (mol : coefficient)
Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³
NaI : 0.0625 : 2 = 0.03125
Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)
moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)
Mass of PbI₂ :
= mol x MW
= 1.36 x 10⁻³ x 461,01 g/mol
= 0.627 g
% yield = 0.55/0.627 x 100% = 87.72%
