Answer:
16.6 g of Al are produced in the reaction of 82.4 g of AlCl₃
Explanation:
Let's see the decomposition reaction:
2AlCl₃ → 2Al + 3Cl₂
2 moles of aluminum chloride decompose to 2 moles of solid Al and 3 moles of chlorine gas.
We determine the moles of salt:
82.4 g . 1mol/ 133.34g = 0.618 moles
Ratio is 2:2. 2 moles of salt, can produce 2 moles of Al
Then, 0.618 moles of salt must produce 0.618 moles of Al.
Let's convert the moles to mass → 0.618 mol . 26.98g /mol = 16.6 g
Answer:
Higher oxidation state metals form stronger bong with ligands
Explanation:
Ligand strength are based on oxidation number, group and its properties
Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).
Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.
Balanced chemical reaction:
Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).
Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).
Answer:
There are 
grams contained in all the seawater in the world.
Explanation:
At first let is determinate the total mass of seawater (
), measured in grams, in the world by definition of density and considering that mass is distributed uniformly:
Where:
- Density of seawater, measured in grams per liters.
- Volume of seawater, measured in liters.
If 
and 
, then:
The total mass of sodium chloride is determined by the following ratio:
Given that and 
, the total mass of sodium chloride in all the seawater in the world is:
There are grams contained in all the seawater in the world.
