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:
the same number of atomic orbitals.
Explanation:
Answer:
inches of mercury (inHg or “Hg) or millibars.
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Answer:
0.76 mole of Fe2S3.
Explanation:
Step 1:
Determination of the number of mole in 449g iron(III)bromide, FeBr3. This is illustrated below:
Mass of FeBr3 = 449g
Molar mass of FeBr3 = 56 + (80x3) = 296g/mol
Mole of FeBr3 =..?
Mole = Mass /Molar Mass
Mole of FeBr3 = 449/296
Mole of FeBr3 = 1.52 moles
Step 2:
The balanced equation for the reaction. This is given below:
2FeBr3 + 3Na2S —> 6NaBr + Fe2S3
Step 3:
Determination of the number of mole of Fe2S3 produced from the reaction of 449g ( i.e 1.52 moles) of FeBr3. This is illustrated below:
From the balanced equation above,
2 moles of FeBr3 reacted to produce 1 mole of Fe2S3.
Therefore, 1.52 moles of FeBr3 will react to produce = (1.52 x 1)/2 = 0.76 mole of Fe2S3.
Therefore, 0.76 mole of Fe2S3 is produced from the reaction.