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neonofarm [45]
3 years ago
6

Which of the following quantum number combinations is not allowed in an atom?

Chemistry
2 answers:
marysya [2.9K]3 years ago
8 0
The fourth combination is not allowed.


The third quantum number, i.e. m subscript l, only may an integer number from -l to + l. Given that l = 0, the only allowed m subscript l is 0.
nika2105 [10]3 years ago
5 0

The answer is n = 1, l = 0, m subscript l = 1

The 4 quantum numbers of the electronic configuration

n: main quantum number. It defines the layer.

n = 1 at layer K

n = 2 at layer L

n = 3 at layer M

And so on…

The energy of the electron is a function of n.

l: secondary or azimuth quantum number

it is an integer that varies from 0 to n-1. It defines the sub-layers s, p, d, f

l = 0 under layer s

l = 1 under layer  p

l = 2 under layer d

l = 3 under layer f

It defines the shape and symmetry of the orbitals (orbital s, p, d etc ...)

Example n = 2 to l = 0 or l = 1 from which 2 sublayers 1s and 2p

m: magnetic quantum number

It takes values ​​between -l and l (including the values ​​of -l and l). At a value of m, correspond to 2l + 1 value of m. It determines the orbital orientation in space (see the shape of the orbitals.

s: quantum number of spin

The quantum number of spin is defined as the kinetic moment (or angular momentum) of the electron. s = -1 / 2 or ½ (2 directions of rotation of the electron on itself)

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Explanation:

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3 years ago
The atomic mass of magnesium is the weighted average of the atomic masses of
Debora [2.8K]

Answer:

2. All the naturally occurring isotopes of Mg.

Explanation:

You want to know the atomic mass of the magnesium you use in the lab. That’s “natural” magnesium. So, you must use the weighted average of all the naturally occurring isotopes in natural Mg.

1. and 3. are <em>wrong</em>. You won’t get the correct mass for natural Mg if you use only the artificial isotopes for your calculation.

4. is <em>wrong</em>. You must use all the naturally occurring isotopes. The two most abundant isotopes of Mg account for only 90 % of the atoms. If you ignore the other 10 %, your calculation will be wrong.

6 0
3 years ago
a sample of helium occupies a volume of 101.2 mL at a pressure of 790 mmHg. at what pressure would the volume be 120 mL?
AveGali [126]

Answer : The final pressure will be, 666.2 mmHg

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure = 790 mmHg

P_2 = final pressure = ?

V_1 = initial volume = 101.2 mL

V_2 = final volume = 120 mL

Now put all the given values in the above equation, we get:

790mmHg\times 101.2mL=P_2\times 120mL

P_2=666.2mmHg

Therefore, the final pressure will be, 666.2 mmHg

6 0
2 years ago
1. NaOH mass of a solution of 200g in which its percentage is 25%. What mass of sulfuric acid solution is needed to completely n
Ede4ka [16]

Answer:

m_{H2SO4 = 61.25 g

m_{Na2SO4} = 88.75 g

Explanation:

m_{NaOH} = \frac{200 . 25 }{100} = 50 g

⇒ n_{NaOH} = \frac{50}{40} = 1.25 (moles)

2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O

   2        :     1           :      1         :    2

 1.25                                                       (moles)

⇒  n_{H2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ m_{H2SO4} = 0.625 × 98 = 61.25 g

    n_{Na2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒m_{Na2SO4} = 0.625 × 142 = 88.75 g

4 0
3 years ago
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