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lubasha [3.4K]
3 years ago
13

The maximum number of orbits in an atoms is____. A) 3 B) 5 C) 7 D) 9

Chemistry
2 answers:
Furkat [3]3 years ago
7 0
I think it's C) seven
Mashcka [7]3 years ago
5 0
The answer///////////b, 5 orbits
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Which model of the atom was used as a result of JJ Thomson's cathode ray
mariarad [96]

Answer:

a. Plum pudding model

Explanation:

The plum pudding model of the atom was proposed by J.J. Thomson. It was the model he derived from his experiment on the gas discharge tube.

J.J Thomson was the first person to discover electrons which he called cathode rays because in the discharge tube, they emanate from the cathode.

  • This led him to suggest the plum pudding model of the atom.
  • The model reflects electrons being surrounded by a volume of negative charges.
6 0
3 years ago
The reason it takes so much heat to melt ice or evaporate water is that extra energy is required to break _______ bonds between
mart [117]

It takes so much heat to melt ice or evaporate water because extra energy is required to break the hydrogen bonds between water molecules.

<h3>Hydrogen bond</h3>

Hydrogen bonding is an electrostatic force of attraction between a hydrogen atom found between a pair of other atoms having a high affinity for electrons.

Hydrogen bonds cause water to be exceptionally attracted to each other creating cohesion.

It takes so much heat to melt ice or evaporate water because extra energy is required to break the hydrogen bonds between water molecules.

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4 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]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

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

  • <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 = 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
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