Which stage is the middle stage of speciation? (15 Points)

katrin2010 [14]

Answer:

₁₁A

Explanation:

Atomic radius

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons.

This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases.

So in given elements consider A₁₁, B₁₂, C₁₃ ans D₁₇ as sodium, magnesium, aluminium and chlorine. This is the third period and as we move form sodium to chlorine atomic radius decreases. That's why sodium has greater size.

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased

7 0

3 years ago

Which refers to the chemical name for NO2?

Anna35 [415]

Answer:

C

Explanation:

6 0

2 years ago

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If three potatoes have a mass of 667 g, what will be the mass of 100 potatoes? A. 200 kg B. 22.2 kg C. 2223 g D. 20.0 kg

adell [148]

Answer: B- 22.2 kg

Explanation: If three potatoes have mass of 667 g that means that each potato weighs 667/3= 222.33 g (approx) so 100 potatoes must be 100*222.33= 22233 g which equals 22.2 kg because 1 g=1000 kg

8 0

3 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

What is a nuclear acid

Ronch [10]

Answer:

Nucleic acid is an important class of macromolecules found in all cells and viruses. Deoxyribonucleic acid (DNA) encodes the information the cell needs to make proteins.

A related type of nucleic acid, called ribonucleic acid (RNA), comes in different molecular forms that participate in protein synthesis.

8 0

2 years ago

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