Ammonium carbonate will form 3 moles of ions.
Methyl alcohol will form 0 moles of ions.
Methane will form 0 moles of ions.
Aluminum sulfite will form 3 moles of ions.
Hydrobromic acid will form 2 moles of ions.
Explanation:
One mole of ammonium carbonate will form 3 moles of ions when dissolved in water.
(NH₄)₂CO₃ (s) + H₂O (l) → 2 NH₄⁺ (aq) + CO₃²⁻ (aq) + H₂O (l)
One mole of methyl alcohol will form 0 moles of ions when dissolved in water.
(actually it form ions because of its acidic behavior but they are at the order of 10⁻⁷ moles, however in the framework of this question we may say that there are none)
One mole of methane will form 0 moles of ions when dissolved in water.
Methane does not react with water (in normal conditions) so will not form ions.
One mole of aluminum sulfite will form 3 moles of ions when dissolved in water.
Al₂SO₃ (s) + H₂O (l) → 2 Al₃⁺ (aq) + SO₃²⁻ (aq) + H₂O (l)
One mole of hydrobromic acid will form 2 moles of ions when dissolved in water.
HBr (l) + H₂O (l) → Br⁻ (aq) + H₃O⁺ (aq)
Learn more about:
solvation of ions
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Answer:
CaO + H₂O ⟶ Ca²⁺ + 2OH⁻
Explanation:
CaO + H₂O ⟶ Ca²⁺ + 2OH⁻
base acid conj. acid conj. base
Given:
7.20 g sample of Al2(SO4)3
Required:
Mass of oxygen
Solution:
Since you are not given a
chemical reaction, just base your solution to the chemical formula given.
Molar mass of Al2(SO4)3 = 342.15 g/mol
7.20 g Al2(SO4)3 (1 mol/342.15g)(3mol O/2 mol Al)(1 mol O2/1/2 mol
O2)(32g O2/1mol O2) = 4.04 g O2
The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4
Evgesh-ka [11]
In NH3 , let oxidation number of N be x
x + (+1)3 = 0
x = -3
In HNO3 , let oxidation number of N be x
1 + x + (-2)3 = 0
x = +5
In NO2 , let oxidation number of N be x
x + (-2)2 = 0
x = +4
