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4 years ago

Mass in grams of 6.25 mol of copper (II) nitrate?

2 answers:

8 0

Cu = 63.546
N= 14.001 g/mol
O= 15.999 g/mol * 3 = 47.997

Copper (II) Nitrate has a MW of 125.544 g/mol

6.25 x 125.544

= 784.65 <--- is your answer, if there were was a multiple choice or not :)

Margaret [11]4 years ago

3 0

Answer:

Mass of 6.25 moles of copper(II) nitrate = 1025 grams

Explanation:

Let us calculate the molar mass of Copper (II) nitrate

Molecular formula = Ca(NO₃)₂

Atomic mass of Ca = 40 g/mol

Atomic mass of N =14 g/mol

Atomic mass of O = 16 g/mol

molar mass of Ca(NO₃)₂ = 40 + (14 X2) +( 6X16) = 164 g/mol

$moles=\frac{mass}{molarmass}$

therefore

mass = moles X molar mass

mass = 6.25 X 164 = 1025 grams

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Answer:

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2p⁴ ║ ↑↓ ║ "↑" ║ ↑

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n = 2,
l = 1,

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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

n = 2, shell number 2,
l = 1, subshell or orbital p,

$m_{l}$ = 0, 2nd "box" in the range -1, 0, 1

$m_{s}$ = +1/2, single electron always has +1/2

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