Answer:
Here's what I get
Explanation:
1. In dilute NaOH
(a) Molecular equation
AlCl₃(aq) + 3NaOH(aq) ⟶ Al(OH)₃(s) + 3NaCl(aq)
(b) Ionic equation
You write the molecular formulas for solids, and you write the soluble ionic substances as ions.
Al³⁺(aq) + 3Cl⁻(aq) + 3Na⁺(aq) + 3OH⁻(aq) ⟶ Al(OH)₃(s) + 3Na⁺(aq) + 3Cl⁻(aq)
(c) Net ionic equation
To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.
Al³⁺(aq) + <u>3Cl⁻(aq)</u> + <u>3Na⁺(aq</u>) + 3OH⁻(aq) ⟶ Al(OH)₃(s) + <u>3Na⁺(aq)</u> + <u>3Cl⁻(aq)
</u>
The net ionic equation is
Al³⁺(aq) + 3OH⁻(aq) ⟶ Al(OH)₃(s)
2. In excess NaOH
The aluminium hydroxide reacts with excess hydroxide to form sodium tetrahydroxoaluminate(III).
(a) Molecular equation
AlCl₃(aq) + 4NaOH(aq) ⟶ NaAl(OH)₄(aq) + 3NaCl(aq)
(b) Ionic equation
Al³⁺(aq) + 3Cl⁻(aq) + 4Na⁺(aq) + 4OH⁻(aq) ⟶ Na⁺Al(OH)₄⁻(aq) + 3Na⁺(aq) + 3Cl⁻(aq)
(c) Net ionic equation
Al³⁺(aq) + 4OH⁻(aq) ⟶ Al(OH)₄⁻(aq)
it's the last one because it's depended on the rock layers
I believe it to be g/mol of Calcium carbonate
because to finde mass...u must have grams(g) as units....
it is the only one that have g as units
as for the first answer the avogadros number gives u the number of atoms in one mole of calcium carbonate....
the second one is based on ... At s.t.p one mole of gas occupies 22.4 dm³⇒to find volume
Answer:
0.29mol/L or 0.29moldm⁻³
Explanation:
Given parameters:
Mass of MgSO₄ = 122g
Volume of solution = 3.5L
Molarity is simply the concentration of substances in a solution.
Molarity = number of moles/ Volume
>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.
Number of moles = mass/ molar mass
Molar mass of MgSO₄:
Atomic masses: Mg = 24g
S = 32g
O = 16g
Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol
= (24 + 32 + 64)g/mol
= 120g/mol
Number of moles = 122/120 = 1.02mol
>>>> From the given number of moles we can evaluate the Molarity using this equation:
Molarity = number of moles/ Volume
Molarity of MgSO₄ = 1.02mol/3.5L
= 0.29mol/L
IL = 1dm³
The Molarity of MgSO₄ = 0.29moldm⁻³
