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3 years ago

What is the theoretical yield of ammonia that can be obtained from the reaction of 17.1 g of H2 and excess N2?

1 answer:

6 0

The theoretical yield is the yield after a certain reaction occurs. So if you are given 17.1 g of H2 and an excess of N2, you should know the balanced equation.

N2 + 3H2 → 2NH3

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17 g/mol

(17.1 g of H2)(1 mol H2/2.02g H2)(2 mol NH3/3mol H2)(17g NH3/1mol NH3) = 95.94g NH3

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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

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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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3 years ago

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victus00 [196]

Answer:

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