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zaharov [31]
3 years ago
11

1 point

Chemistry
1 answer:
lawyer [7]3 years ago
6 0

Answer:

B

Explanation:

you're moving the decimal 8 spots to the left so it can only be B

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Earth orbits the____ in a imaginary plane called______
Montano1993 [528]
Earth orbits the sun in a imaginary plane called ecliptic 
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3 years ago
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Describe an electron cloud. Then, name and explain a particular atomic model that used the concept of the electron cloud.
pochemuha

Electron cloud is the region around the nucleus in an atom where we can locate an electron.

The concept of electron cloud model was introduced by the Schrodinger and Heisenberg. According to this model, it would be difficult to know the position of the electrons in an atom and they are not particles that orbit around the nucleus.  We can only expect the electrons to be present in specific areas called the electron clouds around the nucleus. It is the quantum mechanical model that used the concept of electron clouds. According to the model, the electron cloud or an orbital is a space around the nucleus in an atom where the probability of finding an electron is 90%. It explains that electrons show wave nature. It is difficult to determine the exact position and momentum of an electron in an atom.

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The decomposition of nitrogen dioxide to nitrogen monoxide and oxygen gas is a second order process as suspected in the previous
lesya [120]

Explanation:

For the given reaction 2NO_{2} \rightarrow 2NO + O_{2}

Now, expression for half-life of a second order reaction is as follows.

                  t_{1/2} = \frac{1}{[A_{0}]k}     ....... (1)

Second half life of this reaction will be t_{1/4}. So, expression for this will be as follows.

          t_{1/4} = \frac{1}{k} [\frac{1}{[A]_{f}} - \frac{1}{[A_{0}]}]  ...(2)

where [A]_{f} is the final concentration that is, \frac{[A]_{0}}{4} here and [A]_{i} is the initial concentration.

Hence, putting these values into equation (2) formula as follows.

        t_{1/4} = \frac{1}{k} [\frac{4}{[A]_{0}} - \frac{1}{[A_{0}]}]

                      = \frac{3}{[A_{0}]k}     ...... (3)

Now, dividing equation (3) by equation (1) as follows.

           \frac{t_{1/4}}{t_{1/2}} = \frac{3}{[A_{0}]k} \times [A_{0}]k                  

                                    = 3

or,                       t_{1/4} = 3 t_{1/2}    

Thus, we can conclude that one would expect the second half-life of this reaction to be three times the first half-life of this reaction.

6 0
3 years ago
If 7.84 x 10^7 J of energy is released from a fusion reaction, what amount of mass in kilograms would be lost? Recall that c = 3
docker41 [41]
Use the formula
E=mc^2

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Answer:

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