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3 years ago

Which energy transition may take place in the atom if its electron has absorbed a photon?

Chemistry

1 answer:

Nitella [24]3 years ago

6 0

When an electron absorbs a photon, a "quantum leap" of its energy transition takes place.

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Lesechka [4]

Answer:

Transition elements are elements which have partially filled d-orbitals and form at least one or more stable ions.

6 0

3 years ago

Calculate the mass ratio of S to O in SO, and then calculate the mass ratio of S to O in SO2, given that the mass of S is approx

ziro4ka [17]

Explanation:

From the knowledge of law of multiple proportions,

mass ratio of S to O in SO:

mass of S : mass of O

= 32 : 16

= 32/16

= 2/1

mass ratio of S to O in SO2:

= mass of S : 2 × mass of O

= 32 : 2 × 16

= 32/32

= 1/1

ratio of mass ratio of S to O in SO to mass ratio of S to O in SO2:

= 2/1 ÷ 1/1

= 2

Thus, the S to O mass ratio in SO is twice the S to O mass ratio in SO2.

8 0

3 years ago

29 Points!!

atroni [7]

The last one would be false

3 0

3 years ago

Read 2 more answers

Please help!!

hichkok12 [17]

It’s 65% average that your answer

6 0

2 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

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

where,

$\rho$ = density = $10.28g/cm^3$

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

M = atomic mass of metal (molybdenum) = 95.94 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:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

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 = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

4 0

3 years ago