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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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The molecular geometry of SO 3 is a. tetrahedral b. bent C. octahedral d. trigonal planar e. pyramidal
dybincka [34]

Answer:

The molecular geometry of SO3 is trigonal planar.

Explanation:

Look at the Lewis

5 0
2 years ago
What is the molarity of a solution that contains 125 g nacl in 4.00 l solution?
bogdanovich [222]

Answer:

[NaCl[ = 0.535M

Explanation:

We determine the moles of solute:

125 g . 1 mol/ 58.45 g = 2.14 moles

Molarity (mol/L) → 2.14 mol / 4L → 0.535M

Molarity is a sort of concentration that indicates the moles of solute in 1L of solution

4 0
3 years ago
For the following electron-transfer reaction:
creativ13 [48]

Answer:

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

Explanation:

Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)

In the oxidation half reaction, the oxidation number increases:

Mn changes from 0, in the ground state to Mn²⁺.

The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.

Silver changes from Ag⁺ to Ag.

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

To balance the hole reaction, we need to multiply by 2, the second half reaction:

Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2

2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)  

Now we sum, and we can cancel the electrons:

2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻

4 0
3 years ago
]\ \textless \ br /\ \textgreater \ \huge \boxed{ \red{φµεรƭเσɳ -}} \ \textless \ br /\ \textgreater \ ​ \ \textless \ br /\ \te
azamat

Answer:

Nitrogen fixation is a chemical process by which molecular nitrogen in the air is converted into ammonia or related nitrogenous compounds in soil or aquatic systems. Atmospheric nitrogen is molecular dinitrogen, a relatively nonreactive molecule that is metabolically useless to all but a few microorganisms.

Explanation:

6 0
2 years ago
Read 2 more answers
A particular radioactive nuclide has a half-life of 1000 years. What percentage of an initial population of this nuclide has dec
Delicious77 [7]

Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

3 0
3 years ago
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