PLZ HELP! An electron moved from a lower energy level to a higher energy level. What most likely happened during the transition?

How Many Atoms Are In 356.13 Grams Of Sn

s344n2d4d5 [400]

There are three atoms of Sn (Stannous or Tin) in 356.13 g of Sn.

One atom of Sn has the atomic mass (mₐ) of 118,71u which means:

356.13/118.71=3 atoms of Sn

The mass number (symbol A) also called atomic mass number or nucleon number is the total number of protons and neutrons in an atomic nucleus. It determines the atomic mass of atoms and it is in the periodic table.

8 0

2 years ago

Which laws can be combined to form the ideal gas law?

ahrayia [7]

Answer: Charles's law, Avogadro's law andd Boyle's law.

Charles law states the constant ratio of volume to temperature, at constant pressure. Boyle's law states the constat product of pressure and volumen at constant temperature. Avogadro's law states that equal volumes of gases at the same temperature and pressure have equal number of particles.

So, all those three laws combined state the relation of pressure, volume, temperature and number of particles of a gas, which is what the ideal gas law does: PV = n RT.

4 0

3 years ago

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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 $m_s = (+1/2)$ .