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Mrac [35]
3 years ago
5

What effect does hybridization have on bonds​

Chemistry
1 answer:
marysya [2.9K]3 years ago
7 0
Effect of Hybridisation on Single, Double and Triple Bond Lengths of Carbon.
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The correct answer is c.
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Which of these best describes the particle motion taking place as CO2 gas is exposed to freezing temperatures?
mote1985 [20]
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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
_2. The temperature of a 100. mL sample of gas is increased while the volume it occupies decreases. The
yaroslaw [1]

Answer

the answer is b bro

Explanation:

i did my research and got it right

5 0
3 years ago
2. The density of helium is 1.78 X 104 g/cm. What is this<br> density in Dg/um??
Zepler [3.9K]

Answer:

d=1.78\times 10^{-7}\ Dg/\mu m^3

Explanation:

Given,

The density of Helium is 1.78\times 10^4\ g/cm^3

We need to find the density in Dg/μm

We know that,

1 g = 10 dg

1 cm³ = 10¹² μm³

So,

d=1.78 \times 10^4\ g/cm^3\\\\=1.78 \times 10^4\times \dfrac{10\ dg}{10^{12}\ \mu m^3}\\\\=1.78\times 10^{-7}\ Dg/\mu m^3

So, the density of Helium is equal to 1.78\times 10^{-7}\ Dg/\mu m^3.

4 0
3 years ago
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