Explanation:
1. Explain how groups 1A-8A in the periodic table are organized by their number of valence electrons.
The valence electrons in an atom are the outermost shell electrons. They are the most loosely held electrons in an atom.
Coincidentally, the periodic table of elements divided into vertical groups and horizontal periods can be said to be arranged according to the number of valence electrons.
- Atomic numbers are used to arrange elements on the periodic table.
- Down a group, the number of electronic shell increases. More electrons are added to new energy levels.
- As we move from left to right across a period, the number of electrons in elements increases but electronic shell is the same.
- Down a group electronic shell increases but the number of valence electrons are the same.
- All elements in Group 1A has just one valence electrons, Group 2A has two valence electrons.........Group 8A has eight valence electrons.
- Moving across groups is synonymous to moving from left to right on the periodic table.
- Due to this trend, the periodic table is arranged based on the number of valence electrons.
3. explain how you know the number of valence electrons for each group.
The number of valence electrons in a group is the group number:
Group Number valence electrons
1A 1
2A 2
3A 3
4A 4
5A 5
6A 6
7A 7
8A 8
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Answer:
A 03
Explanation:
jammer as verkeerd
Ek is nog steeds 'n beginner
Great amounts of atomic energy are released when
a _______reaction occurs.
Great amounts of atomic energy
are released when a chemical reaction occurs. The process can be an exothermic reaction
or endothermic reaction depending on the substances involved in the reaction.
Answer:
Explanation:
From the given information:
The concentration of metal ions are:
![[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%3D%20%5Cdfrac%7B0.003474%20%5C%20M%20%5Ctimes%2020.49%20%5C%20mL%7D%7B10.0%20%5C%20mL%7D)
![[Ca^{2+}]=0.007118 \ M](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%3D0.007118%20%5C%20M)
![[Mg^2+] = \dfrac{0.003474 \ M\times (26.23 - 20.49 )mL}{10.0 \ mL}](https://tex.z-dn.net/?f=%5BMg%5E2%2B%5D%20%3D%20%5Cdfrac%7B0.003474%20%5C%20M%5Ctimes%20%2826.23%20%20-%2020.49%20%29mL%7D%7B10.0%20%5C%20mL%7D)
Mass of Ca²⁺ in 2.00 L urine sample is:
= 0.1598 g
Mass of Ca²⁺ = 159.0 mg
Mass of Mg²⁺ in 2.00 L urine sample is:
= 0.3461 g
Mass of Mg²⁺ = 346.1 mg
