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2 years ago

Describe the main difference between the Bohr model of the atom and the Rutherford model.

Chemistry

2 answers:

atroni [7]2 years ago

8 0

Answer:

The main difference between the two models was about the location of the electron in an atom.

Explanation:

Rutherford described that:

The electrons were outside of the nucleus.

They were equal in number to the "positiveness" of the nucleus.

Also, He randomly placed the negative electrons outside the nucleus.

Bohr improved the Rutherford model:

Bohr placed the electrons in distinct energy levels.

Electrons only exist in fixed orbitals (shells) and not anywhere in between.

Each shell has a fixed energy

Rutherford explained the nucleus of an atom, while Bohr was more into the electrons and their energy levels.

Darina [25.2K]2 years ago

7 0

Exact Answer:

In the Rutherford model, electrons can exist in any location outside the nucleus.

In the Bohr model, electrons can exist only in certain energy levels surrounding the atom.

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The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C

Kay [80]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 35 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 37 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

$35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.775\times 100=77.5\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.775)=0.225\times 100=22.5\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

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3 years ago

A wave with low energy will also have high frequencies and long wavelength. short wavelengths and high frequencies. low frequenc

ch4aika [34]

A wave with low energy will also have long wavelengths and low frequencies.

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E = hc/λ
and
E = hf

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3 years ago

Read 2 more answers

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The compound which is ionic is Kh

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3 years ago

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3 years ago

What is the mass of silver in 3.4g of AgNO3?

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3 years ago