Pure substances contain only one type of a. Atoms only. B. Molecules only. C. Atoms or molecules. D. Mixture.

2 answers:

6 0

Pure substances contain atoms only. You cannot break down a pure substance. An element is always uniform all the way through. An element CANNOT be separated into simpler materials except during nuclear reactions.

Nesterboy [21]3 years ago

3 0

A. Atoms only. An element cannot be separated into simpler materials

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Identify the limiting reactant in the reaction of nitrogen and hydrogen to form NH3 if 5.23 g of N2 and 5.52 g of H2 are combine

Marrrta [24]

Answer:

1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

2) The amount (in grams) of excess reactant H₂ = 4.39 g.

Explanation:

Firstly, we should write the balanced equation of the reaction:

N₂ + 3H₂ → 2NH₃.

1) To determine the limiting reactant of the reaction:

From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.

This means that N₂ reacts with H₂ with a ratio of (1:3).

We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation: n = mass / molar mass.

The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.

The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.

From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).

The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).

∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

2) To determine the amount (in grams) of excess reactant of the reaction:

As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.

∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.

4 0

3 years ago

What is the net ion equation and spectator ions for the reaction of (aq) NaC2H3O2 and aq (NH4)2 CO3?

mestny [16]

The reaction is:

(C₂H₃O₂)²⁻ + NH₄⁺ = (NH₄)(C₂H₃O₂).

$Na^+$ and $CO_3^2 -$ are the spectator ions.

<h3>Explanation:</h3>

The reactants in the reaction is sodium acetate and ammonium carbonate. They form the soluble molecule sodium carbonate and slightly soluble ammonium acetate. This is a double displacement reaction.

Spectator ions in the reaction are those ions which are present in the reaction but they actually don't take part in the reaction that's happening and they remain same before and after the reaction. In this reaction, the spectator ions are sodium and carbonate which remain in ionized form in the solution before and after the reaction.