The element silver has an atomic weight of 108 and consists of two stable isotopes silver-107 and silver-109. The isotope silver
1 answer:
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ANSWER:
4 a) Specific elements have more than one oxidation state, demonstrating variable valency.
For example, the following transition metals demonstrate varied valence states: ,
,
, etc.
Normal metals such as also show variable valencies. Certain non-metals are also found to show more than one valence state
4 b) Isotopes are members of a family of an element that all have the same number of protons but different numbers of neutrons.
For example, Carbon-14 is a naturally occurring radioactive isotope of carbon, having six protons and eight neutrons in the nucleus. However, C-14 does not last forever and there will come a time when it loses its extra neutrons and becomes Carbon-12.
5 a) →
5 b) →
5 c) →
(already balanced so don't need to change)
5 d) →
5 e) →
EXPLANATION (IF NEEDED):
1. Write out how many atoms of each element is on the left (reactant side) and right (product side) of the arrow.
2. Start multiplying each side accordingly to try to get atoms of the elements on both sides equal.
EXAMPLE OF BALANCING:
Answer:
We can not swim in 1.00 × 10²⁷ molecules of water
Explanation:
The given number of molecules of water = 1.00 × 10²⁷ molecules
The Avogadro's number, , gives the number of molecules in one mole of a substance
≈ 6.0221409 × 10²³ molecules/mol
Therefore
Therefore, we have;
The number of moles of water present in 1.00 × 10²⁷ molecules, n = (The number of molecules of water) ÷
∴ n = (1.00 × 10²⁷ molecules)/(6.0221409 × 10²³ molecules/mol) = 1,660.53902857 moles
The mass of one mole of water = The molar mass of water = 18.01528 g/mol
The mass, 'm', of water in 1,660.53902857 moles of water is given as follows;
Mass = (The number of moles of the substance) × (The molar mass of the substance)
∴ The mass of the water in the given quantity of water, m = 1,660.53902857 moles × 18.01528 g/mol ≈ 29.9150756 kg.
The density pf water, ρ = 997 kg/m³
Volume = Mass/Density
∴ The volume of the water present in the given quantity of water, v = 29.9150756 kg/(997 kg/m³) ≈ 30.0050909 liters
The volume of the water present in 1.00 × 10²⁷ molecules of water ≈ 30.0 liters
The average volume of a human body = 62 liters
Therefore, we can not swim in the given quantity of 1.00 × 10²⁷ molecules = 30.0 liters water