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PilotLPTM [1.2K]
3 years ago
14

Recycling of aluminum cans is an example of A) increasing entropy is a nonspontaneous process B) decreasing entropy is a nonspon

tanous process C) increasing entropy is a spontaneous process D) decreasing entropy is a spontaeous process.
Chemistry
1 answer:
Helen [10]3 years ago
3 0

Answer:

B) decreasing entropy is a non-spontaneous process

Explanation:

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In a fluorine atom, on which energy level are the valence electrons found?
aivan3 [116]

Answer:

second energy level

Explanation:

Valence electrons are those electrons which are present in outer most orbital of the atom.

This can be easily found through the electronic configuration of atom.

Electronic configuration of F:

F₉ = 1s² 2s² 2p⁵

We can see that the valence electrons are present in second energy level of F atom.

There are seven valence electrons of fluorine.

It is called halogens.

Halogens are very reactive these elements can not be found free in nature. Their boiling points also increases down the group which changes their physical states. i.e fluorine is gas while iodine is solid.

Fluorine:

1. it is yellow in color.

2. it is flammable gas.

3. it is highly corrosive.

4. fluorine has pungent smell.

5. its reactions with all other elements are very vigorous except neon, oxygen, krypton and helium.  

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3 years ago
Bromine, a liquid at room temperature, has a boiling point of 58° C and a melting point of -7.2° C bromine can be classified as
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The answer to the chemistry question above would be pure substance. Bromine is classified as a pure substance because only the element is present and there are no other substances added or mixed with it. It is also not a compound because it does not come with another element. 
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3 years ago
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What is the name given to the electrons in the highest occupied energy level of an atom? -anions -cations -valence electrons -or
Serhud [2]

Answer:

i know the answer the answer is valence electrons.

5 0
3 years ago
Was the first person t propose the idea of moving continents as a scientific hypothesis
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3 years ago
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An experiment with 55 co takes 47.5 hours. at the end of the experiment, 1.90 ng of 55-co remains. if the half-life is 18.0 hour
Andru [333]

Answer:

\boxed{\text{10.7 ng}}

Explanation:

Let A₀ = the original amount of ⁵⁵Co .

The amount remaining after one half-life is ½A₀.

After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.

After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀.

The general formula for the amount remaining is:

A =A₀(½)ⁿ

where n is the number of half-lives

n = t/t_½

Data:

   A = 1.90 ng

    t = 45 h

t_½ = 18.0 h

Calculation:

(a) Calculate n

n = 45/18.0 = 2.5

(b) Calculate A

1.90 = A₀ × (½)^2.5

1.90 = A₀ × 0.178

A₀ = 1.90/0.178 = 10.7 ng

The original mass of ⁵⁵Co was \boxed{\text{10.7 ng}}.

7 0
3 years ago
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