<span>When a gas is cooled, it loses it's energy (the energy associated with thermal motion of molecules</span>
According to the reaction, when 355 ml of 1.65 m Hydrochloric acid solution are combined with too much aluminum, 26.03 g of aluminum chloride is created.
An inorganic substance with the formula AlCl3 is aluminum chloride, also referred to as aluminum trichloride. It takes the form of [Al(H2O)6] hexahydrate. Aluminum chloride is ALCl3, which is composed of six water molecules. Aqueous hydrochloric acid, also referred to as muriatic acid, is a type of hydrochloric acid. It is a colourless solution with an overpowering odor. Strong acid is how it is categorized.
2 Al(s) + 6 HCl is the given reaction. ———-> 2 AlCl3 + 3H2
Using the provided data
The amount of HCl in moles is n= Volume * Concentration = 0.58575 mol (1.65 M * 355 ml)/1000 ml
based on the response,
It takes 6 mols of HCl. To neutralize 2 moles of aluminum, 0.58575 mol of hydrochloric acid must be used.
using the cross-multiplication method, 0.19525 moles are obtained by multiplying 0.58575% by 2/ 6.
Weight of the AlCl3 = 0.1952 moles * 133.340538 g/mol = 26.03 g
Learn more about Hydrochloric acid here
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The number 23.0702 has 6 significant figures.
To answer this, you must have the rules for significant figures,
1. All non-zero numbers ARE significant
2. Zeros between two non-zero digits ARE significant
3. Leading zeros are NOT significant
4. Trailing zeros are ONLY significant if the number has a decimal point
The number 23.0702 has 4 non-zero numbers(rule 1), and both zeros are between two non-zero digits(rule 2). Since all 6 figures in the number 23.0702 are significant, the number 23.0702 has 6 significant figures.
Answer:
Anode: H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻
Cathode: 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq)
E° = 1.60 V
Explanation:
Let's consider the reaction taking place in a galvanic cell.
2 Fe⁺³(aq) + H₂(g) + 2 OH⁻(aq) → 2 Fe⁺²(aq) + 2 H₂O(l)
The corresponding half-reactions are:
Anode (oxidation): H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻ E°red = - 0.83 V
Cathode (reduction): 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq) E°red = 0.77 V
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.77 V - (-0.83 V) = 1.60 V
