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Lapatulllka [165]

3 years ago

10

The value of the atomic mass of an atom is? please help!!

2 answers:

KATRIN_1 [288]3 years ago

4 0

<span> the average mass of all the different isotopes of the atom</span>

nadezda [96]3 years ago

4 0

<span>The value of the atomic mass of an atom is "The average mass of all the isotopes" we also know that as "Molar Mass"

Hope this helps!</span>

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According to Hund's rule of maximum spin multiplicity, how many singly-occupied orbitals are there in the valence shells of the

leva [86]

Answer:

A) carbon - 2

B) cobalt - 3

C) sulfur - 2

D) fluorine - 1

E) titanium - 2

F) germanium - 2

Explanation:

Hund's rule of maximum multiplicity:-

Firstly, every orbital which is present in the sublevel is singly occupied and then the orbital is doubly occupied.

(A) Carbon.

The electronic configuration is -

$1s^22s^22p^2$

Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, Carbon has 2 singly occupied orbitals.

(B) Cobalt.

The electronic configuration is -

$1s^22s^22p^63s^23p^63d^{7}4s^2$

Thus, 4s orbital is fully filled and d orbital can singly filled 5 electrons. Thus, 4 electrons will be paired in 2 orbitals and 3 orbitals will be singly filled in cobalt.

(C) Sulfur.

The electronic configuration is -

$1s^22s^22p^63s^23p^4$

Thus, 3s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, 2 electrons will be paired in 1 orbital and 2 orbitals will be singly filled in sulfur.

D) fluorine

The electronic configuration is -

$1s^22s^22p^5$

Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, 4 electrons will be paired in 2 orbitals and 1 orbital will be singly filled in fluorine.

E) Titanium

The electronic configuration is -

$1s^22s^22p^63s^23p^63d^{2}4s^2$

Thus, 4s orbital is fully filled and d orbital can singly filled 5 electrons. Thus, 2 orbitals will be singly filled in titanium.

F) Germanium

The electronic configuration is -

$1s^22s^22p^63s^23p^63d^{10}4s^24p^2$

Thus, 4s, 3d orbitals are fully filled and p orbital can singly filled 3 electrons. Thus, Germanium has 2 singly occupied orbitals.

4 0

3 years ago

What happens to a hill due to wind blown sand

Wewaii [24]

I'd say that the answer is erosion

3 0

3 years ago

01.03 Transformation of Energy)

Masja [62]

A quantity of property that must be transferred to body to the physical system to perform work

7 0

3 years ago

Read 2 more answers

Aides in the burning of materials. Needed for humans to breathe.

Veseljchak [2.6K]

Answer:

Oxygen.

Explanation:

Oxygen is a chemical element that aides in the burning of materials and it is needed for humans to breathe.

Cellular respiration can be defined as a series of metabolic reactions that typically occur in cells so as to produce energy in the form of adenosine triphosphate (ATP). During cellular respiration, high energy intermediates are created that can then be oxidized to make adenosine triphosphate (ATP). Therefore, the intermediary products are produced at the glycolysis and citric acid cycle stage through the breathing of oxygen used to obtain energy from the food ingested.

Hence, all cells in living organisms require oxygen and glucose to release energy.

3 0

3 years ago

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate

Delicious77 [7]

<u>Answer:</u>

<u>For a:</u> The edge length of the crystal is 533.5 pm

<u>For b:</u> The atomic radius of potassium is 231.01 pm

<u>Explanation:</u>

<u>For a:</u>

To calculate the lattice parameter or edge length of the crystal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $0.855g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal = 39.09 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell = ?

Putting values in above equation, we get:

$0.855=\frac{2\times 39.09}{6.022\times 10^{23}\times (a)^3}\\\\\a=5.335\times 10^{-8}cm=533.5pm$

<u>Conversion factor:</u> $1cm=10^{10}pm$

Hence, the edge length of the crystal is 533.5 pm

<u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 533.5 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 533.5}{4}=231.01pm$

Hence, the atomic radius of potassium is 231.01 pm

4 0

3 years ago