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qaws [65]
3 years ago
15

BRAINLIESTTT ASAP!!! PLEASE HELP ME :)

Chemistry
1 answer:
andrew11 [14]3 years ago
4 0

The Bohr model was a one-dimensional model that used one quantum number to describe the distribution of electrons in the atom. The only information that was important was the size of the orbit, which was described by the n quantum number. Schr�dinger's model allowed the electron to occupy three-dimensional space. It therefore required three coordinates, or three quantum numbers, to describe the orbitals in which electrons can be found.

The three coordinates that come from Schr�dinger's wave equations are the principal (n), angular (l), and magnetic (m) quantum numbers. These quantum numbers describe the size, shape, and orientation in space of the orbitals on an atom.

The principal quantum number (n) describes the size of the orbital. Orbitals for which n = 2 are larger than those for which n = 1, for example. Because they have opposite electrical charges, electrons are attracted to the nucleus of the atom. Energy must therefore be absorbed to excite an electron from an orbital in which the electron is close to the nucleus (n = 1) into an orbital in which it is further from the nucleus (n = 2). The principal quantum number therefore indirectly describes the energy of an orbital.

The angular quantum number (l) describes the shape of the orbital. Orbitals have shapes that are best described as spherical (l = 0), polar (l = 1), or cloverleaf (l = 2). They can even take on more complex shapes as the value of the angular quantum number becomes larger.

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Answer:

1. The electronic configuration of X is: 1s2 2s2 sp6 3s2

2. The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6

3. The formula of the compound form by X and Y is given as: XY

Explanation:

For X to loss two electrons, it means X is a group 2 element. X can be any element in group 2. The electronic configuration of X is:

1s2 2s2 sp6 3s2

To get the electronic configuration of the anion of element Y, let us find the configuration of element Y. This is done as follows:

Y receives two electrons from X to complete its octet. Therefore Y is a group 6 element. The electronic configuration of Y is given below

1s2 2s2 2p4

The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6

The formula of the compound form by X and Y is given below :

X^2+ + Y^2- —> XY

Their valency will cancel out thus forming XY

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How do the properties of a compound compare to the properties of the elements that make it up? A. the properties of the compound
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Calculate the activity coefficients for the following conditions:
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<u>Answer:</u>

<u>For a:</u> The activity coefficient of copper ions is 0.676

<u>For b:</u> The activity coefficient of potassium ions is 0.851

<u>For c:</u> The activity coefficient of potassium ions is 0.794

<u>Explanation:</u>

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}       ........(1)

where,

\gamma_i = activity coefficient of ion

Z_i = charge of the ion

\mu = ionic strength of solution

\alpha _i = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)        ......(2)

where,

C_i = concentration of i-th ions

Z_i = charge of i-th ions

  • <u>For a:</u>

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M

Now, calculating the activity coefficient of Cu^{2+} ion in the solution by using equation 1:

Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{  (known)}\\\mu=0.01M

Putting values in equation 1, we get:

-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676

Hence, the activity coefficient of copper ions is 0.676

  • <u>For b:</u>

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M

Now, calculating the activity coefficient of K^{+} ion in the solution by using equation 1:

Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{  (known)}\\\mu=0.025M

Putting values in equation 1, we get:

-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851

Hence, the activity coefficient of potassium ions is 0.851

  • <u>For c:</u>

We are given:

0.02 M K_2SO_4 solution:

Calculating the ionic strength by using equation 2:

C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M

Now, calculating the activity coefficient of K^{+} ion in the solution by using equation 1:

Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{  (known)}\\\mu=0.06M

Putting values in equation 1, we get:

-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794

Hence, the activity coefficient of potassium ions is 0.794

6 0
3 years ago
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