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>
The subscriot 2 means that in the formula there are two parts of K, and the subscript 1 (implicit) for S, indicates that there is one part of S.
This is, the formula gives the ratio of the elements K and S in the compound, which is:
2 atoms of K : 1 atom of S.
Answer: there are 2 atoms of K and 1 atom of S in a molecule of K2S.
The answer is 0784 the second one
Answer:
Any element with 7 electrons in the outermost shell will have similar properties. Thus other elements in the same column of the periodic table as chlorine will have similar properties. Elements in this column form what is called the Halogen series, which is composed of fluorine, chlorine, bromine, iodine, and astatine
Answer:
14.533 grams of solid precipitate of mercury(II) dichromate will form.
Explanation:
Moles of mercury(II) acetate = 
Moles of sodium dichromate = 
According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :
of mercury(II) acetate
This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.
According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :
of mercury(II) dichromate
Mass of 0.045906 moles of mercury(II) dichromate:
0.045906 mol × 316.59 g/mol = 14.533 g
14.533 grams of solid precipitate of mercury(II) dichromate will form.
