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goldfiish [28.3K]
3 years ago
14

How can an electron cloud be compared with a spinning airplane propeller? Both represent objects that move in a circular motion

around a central core. Exact locations within either cannot be determined at any given moment in time. Either model can be used to picture the shape of an atom.
Chemistry
2 answers:
adelina 88 [10]3 years ago
6 0

The answer is that exact locations within either cannot be determined at any given moment in time.

An electron cloud be compared with a spinning airplane propeller in the manner that in both exact location within either cannot be determined at any given moment in time.

In both electron cloud as well as spinning airplane propeller, there is a probability of finding either but exact location can not be determined.


Serhud [2]3 years ago
5 0

Answer: Option (b) is the correct answer.

Explanation:

It is known that an atom has protons, neutrons and electrons as subatomic particles.

Protons and neutrons are placed inside nucleus of an atom whereas electrons revolve around the nucleus.

Therefore, when a spinning airplane propeller is compared with a spinning electron then it means the propeller also moves in circular motion with a high speed.

Therefore, we cannot tell the exact location of propeller at any given moment in time.

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Answer:

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Explanation:

Step 1: Write the unbalanced chemical equation,

                          C₄H₁₀ + O₂ → CO₂ + H₂O

Step 2: Balance Carbon  Atoms;

As there are 4 carbon  atoms on left hand side and 1 carbon atoms on right hand site therefore, to balance them multiply CO₂ on right hand side by 4 i.e.

                            C₄H₁₀ + O₂ → 4 CO₂ + H₂O

Step 3: Balance Hydrogen Atoms;

There are 10 hydrogen atoms on left hand side and 2 hydrogen atom on right hand site therefore, to balance them multiply H₂O on left hand side by 5 i.e.

                            C₄H₁₀ + O₂ → 4 CO₂ +  5 H₂O

Step 4: Balance Oxygen Atoms:

Now there are 2 oxygen atoms in reactant side and 13 in product side. So, multiply O₂ by 6.5 i.e.

                            C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O

Step 5: Remove fraction coefficients as,

Multiply whole equation by 2 to get rid of fractions i.e.

                            2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

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A chemical reaction of exothermic kind releases energy in the form of heat and light.

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For further reference:

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