Chlorite is ClO2^-1, and potassium is K with a charge of +1. So, we would write is as KClO2, since the charges cancel out with each other.
Distillation; either simple or fractional macroscale depending on how far apart the boiling points are.
Answer:
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Explanation:
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Answer:
Net ionic equation:
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
Explanation:
Chemical equation:
BaCl₂ + Na₂SO₄ → BaSO₄ + NaCl
Balanced Chemical equation:
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)
Ionic equation:
Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)+ 2Na⁺(aq) + 2Cl⁻ (aq)
Net ionic equation:
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
The Cl⁻(aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The BaSO₄ can not be splitted into ions because it is present in solid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
Answer:
-The other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide.
-It is reasonable to exclude iodides and bromides but it is not reasonable to exclude other chlorides
Explanation:
In the qualitative determination of halogen ions, silver nitrate solution(AgNO3) is usually used. Now, various halide ions will give various colours of precipitate when mixed with with silver nitrate. For example, chlorides(Cl-) normally yield a white precipitate, bromides(Br-) normally yield a cream precipitate while iodides (I-) normally yield a yellow precipitate. Thus, all these ions or some of them may be present in the system.
With that being said, if other chlorides are present, they will also yield a white precipitate just like KCl leading to a false positive test for KCl. However, since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. Thus, we can exclude other halides from the tendency to give us a false positive test for KCl but not other chlorides.