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Fittoniya [83]
3 years ago
9

Which of the following is a possible set of quantum numbers for an electron (n,l,m0,ms) ?

Chemistry
1 answer:
Mashutka [201]3 years ago
5 0

Answer is: (3, 2, 0, -1/2).

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=3:  

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

l = 0, 1, 2.  

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 = -2, -1, 0, +1, +2.

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.  

(1, 1, 0, +1/2)  is not correct because orbital quantum number cannot be l = 1 for n = 1.

(2, 1, 2, +1/2)  is not correct because magnetic quantum number cannot be ml = 2 for orbital quantum number l = 1.

(3, -2, 1, -1/2) is not correct because orbital quantum number cannot be l = -2 for principal quantum number n = 3.

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What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa
almond37 [142]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of stomach acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL

Putting values in above equation, we get:

1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL

Hence, the volume of HCl neutralized is 1.25 mL

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3 years ago
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