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1 answer:

6 0

The correct answer is C because a Controlled Experiment tests only one factor at a time and consists of a control group and one or more experimental groups.

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A solution is prepared by dissolving a

goldfiish [28.3K]

Answer:

119g

Explanation:

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2.38% of 500 =

119 g

7 0

3 years ago

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Use the periodic table to identify the element with the electron configuration 1s²2s²2p⁴. Write its orbital diagram, and give th

natta225 [31]

Answer:

1. Orbital diagram

2p⁴ ║ ↑↓ ║ "↑" ║ ↑

2s² ║ ↑↓ ║

1s² ║ ↑↓ ║

2. Quantum numbers

n = 2,
l = 1,

$m_{l}$ = 0,

$m_{s}$ = +1/2

Explanation:

The fill in rule is:

Follow shell number: from the inner most shell to the outer most shell, our case from shell 1 to 2

Follow the The Aufbau principle, 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
Hunds' rule: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).

So, the orbital diagram of given element is as below and the sixth electron is marked between " "

2p⁴ ║ ↑↓ ║ "↑" ║ ↑

2s² ║ ↑↓ ║

1s² ║ ↑↓ ║

The quantum number of an electron consists of four number:

n (shell number, - 1, 2, 3...)
l (subshell number or orbital number, 0 - orbital s, 1 - orbital p, 2 - orbital d...)
$m_{l}$ (orbital energy, or "which box the electron is in"). For example, orbital p (l = 1) has 3 "boxes", it was number from -1, 0, 1. Orbital d (l = 2) has 5 "boxes", numbered -2, -1, 0, 1, 2
$m_{s}$ (spin of electron), either -1/2 or +1/2

In our case, the electron marked with " " has quantum number