Answer:
!atoms in the nitrogen family.. have 5 valence electrons. They tend to share electrons when they bond. Other elements in this family are phosphorus, arsenic, antimony, and bismuth.
Answer:
the answer would be Halogens
Answer:
The answer to your question is: D. Fractional distillation
Explanation:
A. Column chromatography this method is use to separate mixtures base in its capacity to be absorbed.
B. Fractional crystallization this method use the solubility of solutes in a solvent.
C. Simple distillation is a method to separate 2 liquids based on the difference in their boiling points.
D. Fractional distillation is a method used to separate a mixture of liquids which have their boiling points very close.
E. Paper chromatography is a method to separate a mixture based on their different rate of migration on a paper.
Answer:
The last option:
- NH₃ (aq) + H⁺ (aq) → NH₄⁺ (aq)
Explanation:
1) Word equation
- Aqueous ammonia + nitric acid → aqueous ammonium nitrate
2) Chemical (molecular) equation
- NH₃ (aq) + HNO₃ (aq) → NH₄ NO₃
3) Ionization reactions
Write the dissociation of the soluble ionic compounds:
4) Total ionic equation:
- NH₃ (aq) + H⁺ (aq) + NO₃⁻ (aq) → NH₄⁺ (aq) + NO₃⁻ (aq)
5) Net ionic equation
You must cancel the spectator ions, which are those ions that are repeated in both reactant and product sides, i.e. NO₃⁻. They are name spectator because they do not participate (change) during the reaction.
- NH₃ (aq) + H⁺ (aq) → NH₄⁺ (aq)
And that is the last choice of the list.
Answer:
SO₃²⁻ is the reducing agent and Cr₂O₇²⁻ is the oxidizing agent.
Explanation:
Oxidation reaction:
3SO₃²⁻ (aq) + 3H₂O (l) → 3SO₄²⁻ (aq) + 6H⁺ (aq) + 6e⁻
Reduction reaction:
Cr₂O₇²⁻ (aq) + 14H⁺ (aq) + 6e⁻ → 2Cr ³⁺ (aq) + 7H₂O (l)
Now, adding the oxidation and the reduction reactions we get the full net reaction:
Cr₂O₇²⁻ (aq) + 3SO₃²⁻ (aq) + 8H⁺ (aq) → 2Cr ³⁺ (aq) + 3SO₄²⁻ (aq) + 4H₂O(l)
Since, the S in SO₃²⁻, present in the +4 oxidation state is oxidized to +6 oxidation state in SO₄²⁻, by the loss of 2e⁻.
<u>Therefore, SO₃²⁻ is the reducing agent. </u>
And, the Cr in Cr₂O₇²⁻, present in the +6 oxidation state is getting reduced to +3 oxidation state, Cr ³⁺, by the gain of 6e⁻.
<u>Therefore, Cr₂O₇²⁻ is the oxidizing agent.</u>
