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butalik [34]
3 years ago
11

On the periodic table, what is a group? For the main groups, what characteristic

Chemistry
1 answer:
rjkz [21]3 years ago
4 0
<h3>Answer: Mendeleev’s Periodic Table was built by listing the known elements in order of increasing atomic mass and grouping the elements by their properties (such as reactivity and formula of oxides). In one or two cases, he discovered that he had to put a lower atomic mass element after a higher atomic mass element because the table which went strictly by increasing atomic mass produced groups with inconsistent properties. He also left blanks in the table.</h3>

In modern times, we know that the elements in a given group (also known as a family or column) have the same number of valence electrons, with ground state configurations having a consistent pattern, for instance:

F 1s2 2s2 2p5, Cl 1s2 2s2 2p6 3s2 3p5, Br 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5, etc.

Of course, the ground state configurations play a major role in determining how an element reacts and what compounds it forms. Thus, the elements in a given group have similar properties.

The structure of the table also predicts how the properties vary. For instance, atomic radius increases down a group as does metallic character.

<h3>Hope this helps have a awesome day/night✨</h3>

Explanation:

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When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction: 2 Al (s) + 3 Cl2 (g) --&gt; 2 AlCl3 (s) c) How many moles of the e
sineoko [7]

Answer:

The answer to your question is 1.49 mol of Cl₂  

Explanation:

Data

mass of Al = 40.5 g

mass of Cl₂ = 212.7 g

moles of excess reactant = ?

- Balanced chemical reaction

                2 Al(s)  +  3Cl₂(g)  ⇒   2AlCl₃

Process

a) Calculate the molar mass of the reactants  

molar mass of Al = 2 x 26.98 = 53.96 g

molar mass of Cl₂ = 6 x 35.45 = 212.7 g

b) Calculate the theoretical proportion  Al/Cl₂ = 53.96/212.7 = 0.254

   Calculate the experimental proportion Al/Cl₂ = 40.5/212.7 = 0.19

As the experimental proportion is lower than the theoretical proportion we conclude that the limiting reactant is Aluminum.

c) Calculate the grams of excess reactant

                    53.96 g of Al ------------------ 212.7 g of Cl₂

                     40.5 g of Al -------------------  x

                      x = (40.5 x 212.7) / 53.96

                      x = 8614.35 / 53.96

                      x = 159.64 g of Cl₂

Excess Cl₂ = 212.7 - 159.64

                  = 53.057 g

d) Calculate the moles of Cl

                       35.45 g of Cl ----------------- 1 mol

                       53.057 g of Cl ---------------  x

                        x = (53.057 x 1)/35.45

                       x = 1.49 mol of Cl₂                    

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3 years ago
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Answer:

m=dxv

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Explanation:

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Answer:

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