Highlight the basic points in Lewis and Langmuir theory of electrovalency
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The Lewis and Langmuir theory of electrovalency (and as well as Kossel's) is dealing with Ionic bonds.
Lewis: electron-pair sharing, octet rule, Lewis Symbols or StructureLangmuir: introduced term "covalent" bond, and popularized Lewis's ideas
The Lewis-Langmuir electron-pair or covalent bond is referred as the homopolar bond, where the complete transfer of electrons give rise to ionic, or electrovalent bond (1) through attraction of opposite charges.
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Is this method CORRECT
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Answer:
yes
Explanation:
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Atom is the most basic unit of matter. They are the general term used to describe pieces of matter. But a different kind of atom makes up an element. A combination of atom of different elements creates a molecule. For example, you have water; it has a chemical formula of H2O. If you separate H from O, you have two atoms. Two atoms for hydrogen and one atom for oxygen. However, they are of different elements. And when you combine them to form H2O, you create a molecule.
(2) absorb protons (4) oxidize
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Answer: The nucleus of radium-226 is unstable and hence undergoes decay.
Explanation: Radium has many isotopes. One of them is which has 88 protons and 138 neutrons. It is a radioactive isotope and undergoes decay process.
This isotope undergoes alpha - decay and produces Radon-222 isotope.
Equation for alpha - decay follows:
Hence, the nucleus of radium-226 undergoes decay process.
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The mass of water that will be produced if 10.54 g of H₂ react with 95.10 g of O₂ is 94.86 g
Balanced equation
2H₂ + O₂ —> 2H₂O
Molar mass of H₂ = 2 × 1 = 2 g/mol
Mass of H₂ from the balanced equation = 2 × 2 = 4 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ from the balanced equation = 1 × 32 = 32 g
Molar mass of H₂O = (2×1) + 16 = 18 g/mol
Mass of H₂O from the balanced equation = 2 ×18 = 36 g
SUMMARY
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂ to produce 36 g of H₂O
How to determine the limiting reactant
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂
Therefore,
10.54 g of H₂ will react with = (10.54 × 32) / 4 = 84.32 g of O₂
From the calculation made above, we can see that only 84.32 g out of 95.10 g of O₂ given, is needed to react completely with 10.54 g of H₂.
Therefore, H₂ is the limiting reactant.
How to determine the mass of water produced
From the balanced equation above,
4 g of H₂ reacted to produce 36 g of H₂O
Therefore,
10.54 g of H₂ will react to produce = (10.54 × 36) / 4 = 94.86 g of H₂O
Thus, 94.86 g of H₂O were obtained from the reaction.
Learn more about stoichiometry:
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If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:
360 g : 1 L = x g : 30 mL
Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = x g : 30 mL
Now, crossing the products:
x · 1,000 mL = 360 g · 30 mL
x · 1,000 mL = 10,800 g mL
x = 10,800 g ÷ 1,000
x = 10.8 g
So, from 30 mL mixture, 10.8 g of NaCl could be extracted.
Let's calculate the same for 10 mL water instead of 30 mL.
360 g : 1 L = x g : 10 mL
Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = x g : 10 mL
Now, crossing the products:
x · 1,000 mL = 360 g · 10 mL
x · 1,000 mL = 3,600 g mL
x = 3,600 g ÷ 1,000
x = 3.6 g
So, from 10 mL mixture, 3.6 g of NaCl could be extracted.
Now, let's compare:
If from 30 mL mixture, 10.8 g of NaCl could be extracted and from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:
3.6/10.8 = 1/3
Therefore, it will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.