Explanation:
Valence electrons is the number of electrons in the outermost shell of an atom. For the main group elements, the group number represents the number of valence electrons.
Beryllium
This belongs to the group 2 hence, it has valency of 2.
Nitrogen
This belongs to group 5, hence it has valency of 5.
Oxygen
This belongs to group 6, hence it has valency of 6.
Fluorine
This belongs to group 7, hence it has valency of 7.
Magnesium
This belongs to group 2, hence it has valency of 2.
Phosphorus
This belongs to group 5, hence it has valency of 5
Sulphur
This belongs to group 6, it has valency of 6
Chlorine
This belongs to group 7, it has valency of 7
The amount of water that will be produced is 50.36 grams
<h3>Stoichiometric problems</h3>
The metabolism of glucose is represented by the following equation:
The mole ratio of glucose metabolized to the water produced is 1:6.
Mole of 84.0 g glucose = 84/180.156 = 0.4662 moles
Equivalent mole of water = 0.4662 x 6 = 2.7975 moles
Mass of 2.7975 moles water = 2.7975 x 18 = 50.36 grams
More on stoichiometric problems can be found here: brainly.com/question/14465605
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Answer:
44.91% of Oxygen in Iron (III) hydroxide
Explanation:
To solve this question we must find the molar mass of Fe(OH)3 and the molar mass of the oxygen in this molecule. Percent composition will be:
<em>Molar mass Oxygen / molar mass Fe(OH)3 * 100</em>
<em />
<em>Molar mass Fe(OH)3 and oxygen:</em>
1Fe = 55.845g/mol*1 = 55.845
3O = 16.00g/mol*3 = 48.00 - Molar mass of Oxygen
3H = 1.008g/mol*3 = 3.024
55.845 + 48.00 + 3.024 =
106.869g/mol is molar mass of Fe(OH)3
% Composition of oxygen is:
48.00g/mol / 106.869g/mol * 100 =
<h3>44.91% of Oxygen in Iron (III) hydroxide</h3>
Answer:
the raising atomic number
Explanation:
Elements are listed on the periodic table according to their atomic number.
