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guajiro [1.7K]
3 years ago
5

An electron moved from a lower energy level to a higher energy level. What most likely happened during the transition?

Chemistry
2 answers:
hichkok12 [17]3 years ago
6 0

Answer:

The correct answer is statement that a fixed amount of energy was absorbed.

Explanation:

In an atom, the electrons prevail in distinct energy levels. The electrons present in an atom are confined to a particular level of energy, which is located at different distances from the nucleus. The electrons, which are in the highest energy level are known as valence electrons. Inside every energy level lies a space volume where particular electrons are situated.  

The energy is absorbed by the atom when an electron moves to a higher energy level from a lower energy level, and the energy is discharged when an electron moves to a lower energy level from a higher energy level in the form of light.  

nekit [7.7K]3 years ago
5 0

Answer:

Amount of energy was released

Explanation:

Because when energy is release from the shell it makes the electron lose and than go to higher level of energy form

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Answer:

When carbon is burned in air, it forms carbon dioxide gas and releases a large amount of heat and some light:

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a 25 ml volume of a sodium hydroxide solution requires 19.6 mL of a 0.189 M HCl acid for neutralization. A 10 mL volume of phosp
Drupady [299]

Answer: 0.172 M

Explanation:

a) To calculate theconcentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=0.189M\\V_1=19.6mL\\n_2=1\\M_2=?\\V_2=25mL

Putting values in above equation, we get:

1\times 0.189\times 19.6=1\times M_2\times 25\\\\M_2=0.148

b) To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_3PO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=3\\M_1=?\\V_1=10mL\\n_2=1\\M_2=0.148\\V_2=34.9mL

Putting values in above equation, we get:

3\times M_1\times 10=1\times 0.148\times 34.9\\\\M_1=0.172M

The concentration of the phosphoric acid solution is 1.172 M

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3 years ago
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