I would be difficult to remove an electron from a Noble or Inert Gas (also known as the group 8 or 0 elements). This is because they all have filled outermost shells and as such the outermost shell would be held tightly to the nucleus and as such make it difficult to remove. Examples Helium, Neon, Argon, Xenon, Krypton and Radon
The volume (in cm³) gained by a person who gains 11.8 lb of fat is 5830.49 cm³
<h3>What is density? </h3>
The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as
Density = mass / volume
<h3>How to convert pounds to grams </h3>
1 lb = 453.592 g
Therefore,
11.8 lb = 11.8 × 453.592
11.8 lb = 5352.3856 g
<h3>How to determine the volume </h3>
- Mass = 5352.3856 g
- Density = 0.918 g/cm³
- Volume =?
Volume = mass / density
Volume = 5352.3856 / 0.918
Volume = 5830.49 cm³
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Answer:
The 2 would be placed in front of the reactant Na and in front of the product NaCl
2Na + Cl2 = 2NaCl
Explanation:
This is because the ratio of elements needs to be balanced on both sides.
On the reactants side, there are 2 Na molecules and 2 Cl molecules
On the products side, there are 2 Na molecules and 2 Cl molecules
So, now the equation is balanced
Answer:
1.35 g
Explanation:
Data Given:
mass of Potassium Permagnate (KMnO₄) = 3.34 g
Mass of Oxygen: ?
Solution:
First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)
So,
Molar Mass of KMnO₄ = 39 + 55 + 4(16)
Molar Mass of KMnO₄ = 158 g/mol
Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).
Mass contributed by Oxygen (O) = 4 (16) = 64 g
Since the percentage of compound is 100
So,
Percent of Oxygen (O) = 64 / 158 x 100
Percent of Oxygen (O) = 40.5 %
It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.
So,
for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be
mass of Oxygen (O) = 0.405 x 3.34 g
mass of Oxygen (O) = 1.35 g