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
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Step 1:
Write molecular Formula,
Na₅P₃O₁₀
Step 2:
Assign Oxidation number to Na (which is +1) and O (which is -2), and X (unknown) for P.
(+1)₅ + (X)₃ + (-2)₁₀ = 0
Zero means overall charge on molecule is zero (Neutral).
Step 3:
Now solve for X,
(+5) + (X ÷ 3) + (-20) = 0
X ÷ 3 = -5 + 20
X ÷ 3 = +15
X = +15 ÷ 3
X = +5
Result:
Oxidation number of P is +5.
Answer:
V₂ = 111.3 mL
Explanation:
Given data:
Initial volume of gas = 50.0 mL
Initial temperature = standard = 273.15 K
Final volume = ?
Final temperature = 335 °C (335+273.15 = 608.15 K)
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 50.0 mL ×608.15 K / 273.15 k
V₂ = 30407.5 mL.K / 273.15 K
V₂ = 111.3 mL
Answer:
Right
Explanation:
The given parameters are;
The soluble sulfate formed by the metal, M = M₂SO₄
In the galvanic cell formed by the metal M, we have;
The concentration of M₂SO₄ in the left half cell = 50.0 mM = 0.05 M
The concentration of M₂SO₄ in the right half cell = 5.00 M
In the galvanic cell, the metal 'M' will be dissolved into the solution with lower concentration as M²⁺ which is the left half cell, making the cell negative and the solution more concentrated
In the right half cell, the metal 'M²⁺' in the solution will be plated unto the electrode making the solution less concentrated and the electrode in the right half cell will be the positive electrode
Therefore;
The electrode which will be positive is the electrode in the right half cell.
Hello.
<span>HI + NaOH --> NaI + H2O
2HClO4 + Sr(OH)2 --> Sr(ClO4)2 + 2H2O
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