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3 years ago

Explain why the quantum number set (2, 2, -1, -½) is not possible for an electron in an atom.

Chemistry

1 answer:

rosijanka [135]3 years ago

8 0

Answer is: it is not possible, because orbital (azimunthal) quantum number cannot be 2.

The principal quantum number (n) is one of four quantum numbers which are assigned to each electron in an atom to describe that electron's state.

For principal quantum number n=2:

1) azimuthal quantum number (l) can be l = 0...n-1:

l = 0, 1.

The azimuthal quantum number determines its orbital angular momentum and describes the shape of the orbital.

2) magnetic quantum number (ml) can be ml = -l...+l.

ml = -1, 0, +1.

Magnetic quantum number specify orientation of electrons in magnetic field and number of electron states (orbitals) in subshells.

3) the spin quantum number (ms), is the spin of the electron.

ms = +1/2, -1/2.

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Answer: The mass percent of potassium bromide in the mixture is 9.996%

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To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

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Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

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