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lukranit [14]
3 years ago
10

The worst feeling is when you are hurting so badly inside but not a single tear comes out.

Chemistry
1 answer:
guapka [62]3 years ago
8 0

Answer:

chemistry wow

Explanation:

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Find the quantinum numbers n,m,l,s for the last of potassium layer pleasee help explain correctly all
Fantom [35]

Answer:

Quantum numbers of the outermost electron in potassium:

  • n = 4.
  • l  = 1.
  • m_l = 0.
  • Either m_s = 1/2.

Explanation:

Refer to the electron configuration of a potassium atom. The outermost electron in a ground-state potassium atom is in the 4s orbital (fourth s orbital.)

The quantum number n (the principal quantum number) specifies the main energy shell of an electron. This electron is in the fourth main energy shell (as seen in the number four in the orbital.) Hence, n = 4 for this electron.

The quantum number l (the angular momentum quantum number) specifies the shape (s, p, d, etc.) of an electron. l = 1 for s\! orbitals (such as the one that contains this electron.

Quantum numbers n and l specify the shape of an orbital. On the other hand, the magnetic quantum number m_l specifies the orientation of these orbitals in space.

However, s orbitals are spherical. Regardless of the value of n, the only possible m_l value for electrons in s\! orbitals is m_l = 0.

The spin quantum number m_s distinguishes between the two electrons in an orbital. The two possible values of m_s \! are (+1/2) and (-1/2). Typically, the first electron in an orbital is assigned an upward (\uparrow) spin, which corresponds to m_s = (+1/2).

5 0
3 years ago
Name 2 parts of a microscope
Airida [17]
Two parts are stage and coarse focus
6 0
3 years ago
0.65743 In scientific notation
Goryan [66]
202827.0000 is the answer I think idrk
6 0
3 years ago
Which phase of matter is distinguished by a superheated gas that emits light?
mariarad [96]

Answer:

I think it is Plasma

Explanation:

7 0
3 years ago
What is the half-life of an isotope that decays to 6.25% of its original activity in 18.9 hours?
inessss [21]
Radioactive material obeys 1st order decay kinetics,
For 1st order reaction, we have 
k = \frac{2.303}{t}Xlog \frac{\text{initial conc.}}{\text{final conc.}}
where, k = rate constant of reaction

Given: Initial conc. 100, Final conc. = 6.25, t = 18.9 hours

∴ k = \frac{2.303}{18.9} X log \frac{100}{6.25} = 0.1467 hours^(-1)

Now, for 1st order reactions: half life = \frac{0.693}{k} =  \frac{0.693}{0.1467} = 4.723 hours.


8 0
3 years ago
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