Answer:
(a) Alkali metals: Francium (Fr)
(b) Chalcogens: Polonium (Po)
(c) Noble gases: Radon (Rn)
(d) Alkaline earth metals: Radium (Ra)
Explanation:
In the periodic table, the atomic mass increases down the group. Therefore, the last element of a group is the heaviest element of the group.
(a) alkali metals: The chemical elements that are present in group 1 of the periodic table, except hydrogen.
<u>The heaviest member of this group is francium (Fr)</u>
(b) chalcogens: The chemical elements that are present in group 16 of the periodic table
<u>The heaviest member of this group is polonium (Po)</u>
(c) noble gases: The chemical elements that are present in group 18 of the periodic table
<u>The heaviest member of this group is radon (Rn)</u>
(d) alkaline earth metals: The chemical elements that are present in group 2 of the periodic table.
<u>The heaviest member of this group is radium (Ra)</u>
Answer is: 25,06 kJ of energy must be added to a 75 g block of ice.
ΔHfusion(H₂O) = 6,01 kJ/mol.
T(H₂O) = 0°C.
m(H₂O) = 75 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 75 g ÷ 18 g/mol.
n(H₂O) = 4,17 mol.
Q = ΔHfusion(H₂O) · n(H₂O)
Q = 6,01 kJ/mol · 4,17 mol
Q = 25,06 kJ.
The number of atoms of gold in the pure ring are 7.18 × 10²² atoms.
<h3>HOW TO CALCULATE NUMBER OF ATOMS?</h3>
The number of atoms in a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.
The number of moles in the gold (Au) can be calculated by dividing the mass of gold by its molar mass (196.97g/mol).
no. of moles = 23.5g ÷ 196.97g/mol
no. of moles = 0.119mol
Number of atoms in Au = 0.119 × 6.02 × 10²³
no. of atoms = 7.18 × 10²² atoms.
Therefore, the number of atoms of gold in the pure ring are 7.18 × 10²² atoms.
Learn more about number of atoms at: brainly.com/question/15959704
Answer:
(a) a = 5.08x10⁻⁸ cm
(b) r = 179.6 pm
Explanation:
(a) The lattice parameter "a" can be calculated using the following equation:
<em>where ρ: is the density of Th = 11.72 g/cm³, N° atoms/cell = 4, m: is the atomic weight of Th = 232 g/mol, Vc: is the unit cell volume = a³, and </em>
<em>: is the Avogadro constant = 6.023x10²³ atoms/mol. </em>
Hence the lattice parameter is:

![a = \sqrt[3]{1.32 \cdot 10^{-22} cm^{3}} = 5.08 \cdot 10^{-8} cm](https://tex.z-dn.net/?f=%20a%20%3D%20%5Csqrt%5B3%5D%7B1.32%20%5Ccdot%2010%5E%7B-22%7D%20cm%5E%7B3%7D%7D%20%3D%205.08%20%5Ccdot%2010%5E%7B-8%7D%20cm%20)
(b) We know that the lattice parameter of a FCC structure is:

<em>where r: is the atomic radius of Th</em>
Hence, the atomic radius of Th is:
I hope it helps you!