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Nataliya [291]
3 years ago
12

A chemist determines by measurements that moles of nitrogen gas participate in a chemical reaction. Calculate the mass of nitrog

en gas that participates.
Chemistry
1 answer:
Likurg_2 [28]3 years ago
4 0

Answer:

0.56 g

Explanation:

<em>A chemist determines by measurements that 0.020 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of nitrogen gas that participates.</em>

Step 1: Given data

Moles of nitrogen gas (n): 0.020 mol

Step 2: Calculate the molar mass (M) of nitrogen gas

Molecular nitrogen is a gas formed by diatomic molecules, whose chemical formula is N₂. Its molar mass is:

M(N₂) = 2 × M(N) = 2 × 14.01 g/mol = 28.02 g/mol

Step 3: Calculate the mass (m) corresponding to 0 0.020 moles of nitrogen gas

We will use the following expression.

m = n × M

m = 0.020 mol × 28.02 g/mol

m = 0.56 g

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We must generate this reaction rom the equations given.

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(4) Li(g) ⟶Li⁺(g) +e⁻                   IE₁         =   520 kJ·mol⁻¹

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Now, we put these equations together to get the lattice energy.

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The lattice energy of LiCl is \boxed{\textbf{-862 kJ/mol}}.

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<h3>Explanation</h3>

Find the two elements on a periodic table:

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Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.

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A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:

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The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each \text{O}^{2-} ion is twice that on a \text{K}^{+} ion. Each \text{K}^{+} would pair up with two \text{O}^{2-}. Hence the subscript in the formula: \text{K}_{\bf 2}\text{O}.

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Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide \text{CO}_2, the prefix di- indicates that there are two oxygen atoms in the formula \text{CO}_2. However, there's no need for prefix in ionic compounds such as \text{K}_2\text{O}.

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