Hydrogen is usually –1. This is INCORRECT. The oxidation number for H is +1.
Oxygen is usually –2. This is CORRECT.
A pure group 1 element is +1. This is INCORRECT. It does not follow. This will depend on the other elements and the overall charge.
A monatomic ion is 0. This is INCORRECT. Diatomic ion is 0.
Answer:
% composition O = 19.9%
% composition Cu = 80.1%
Explanation:
Given data:
Total mass of compound = 3.12 g
Mass of copper = 2.50 g
Mass of oxygen = 3.12 - 2.50 = 0.62 g
% composition = ?
Solution:
Formula:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition Cu = (2.50 g / 3.12 g)×100
% composition Cu = 0.80 ×100
% composition Cu = 80.1%
For oxygen:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition O = (0.62 g / 3.12 g)×100
% composition O = 0.199 ×100
% composition O = 19.9%
Answer:
18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury
Explanation:
Mercury oxide has molar mass of 216.6 g/ mol. It gas a molecular formula of HgO.
The decomposition of mercury oxide is given by the chemical equation below:
2HgO ----> 2Hg + O₂
2 moles of HgO decomposes to produce 1 mole of Hg
2 moles of HgO has a mass of 433.2 g
433.2 g of HgO produces 216.6 g of Hg
18.0 of HgO will produce 18 × 216.6/433.2 g of Hg = 9.0 g of Hg
Therefore, 18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury
The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
<h3>How to calculate empirical formula?</h3>
The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.
The empirical formula of the given compound can be calculated as follows:
- Hafnium = 55.7% = 55.7g
- Chlorine = 44.3% = 44.3g
First, we convert mass values to moles by dividing by the molar mass of each element
- Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
- Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol
Next, we divide each mole value by the smallest
- Hafnium = 0.312 ÷ 0.312 = 1
- Chlorine = 1.25 ÷ 0.312 = 4
Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
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