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Setler [38]
2 years ago
14

Please help!!! due tomorrow and i’m really stuck!! thank you!

Chemistry
1 answer:
Zarrin [17]2 years ago
7 0
Check the image for answers and explanations.

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Analysis of an ore of calcium shows that it contains 13.61 g calcium and 21.77 g oxygen in a 46.28-g sample. Calculate the perce
kondor19780726 [428]

Answer:

29.41% of Calcium and 47.04% of Oxygen

Explanation:

The percent composition of an atom in a molecule is defined as 100 times the ratio between the mass of the atom and the mass of the molecule.

The mass of the molecule of the problem (Ore) is 46.28g. That means the percent composition of Calcium is:

13.61g / 46.28g * 100 = 29.41% of Calcium

And percent composition of Oxygen is:

21.77g / 46.28g * 100 = 47.04% of Oxygen

5 0
3 years ago
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).

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