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diamong [38]
3 years ago
5

What were the limitations of Dobereiner's classifcation?​

Chemistry
2 answers:
yawa3891 [41]3 years ago
5 0

Answer:

Dobereiner could find only three triads; . i.e total of 9 elements only. However the total number of elements were more than that of those encompassed in Dobereiner's Triad

Explanation:

Serga [27]3 years ago
4 0

Answer:

Dobereiner could find only three triads; . i.e total of 9 elements only. However the total number of elements were more than that of those encompassed in Dobereiner's Triad. Thus, Dobereiner's could not be classify most of the elements known at that time

Explanation:

This work is not my own.... copied from google but hope this helps you.

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Be sure to answer all parts. The balanced equation for the reaction of aluminum metal and chlorine gas is 2Al(s) + 3Cl2(g) → 2Al
Arada [10]

Answer: a) Cl_2 is the limiting reagent

b)  0.27 g of AlCl_3 will be produced.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Al=\frac{0.80g}{27g/mol}=0.030moles

\text{Moles of} Cl_2=\frac{0.23g}{71g/mol}=0.003moles

2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)

According to stoichiometry :

3 moles of Cl_2 require = 2 moles of Al

Thus 0.003 moles of Cl_2 will require=\frac{2}{3}\times 0.003=0.002moles  of Al

Thus Cl_2 is the limiting reagent as it limits the formation of product and Al is the excess reagent.

b) As 3 moles of Cl_2 give = 2 moles of AlCl_3

Thus 0.003 moles of Cl_2 give =\frac{2}{3}\times 0.003=0.002moles  of AlCl_3

Mass of AlCl_3=moles\times {\text {Molar mass}}=0.002moles\times 133g/mol=0.27g

Thus 0.27 g of AlCl_3 will be produced.

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How many electrons are needed in the outer energy levels of most atoms to be chemically stable?
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The air in a 4.00 L tank has a pressure of 1.20 atm . What is the final pressure, in atmospheres, when the air is placed in tank
stellarik [79]

Answer:

P₂ = 1.92 atm

Explanation:

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P₁V₁= P₂V₂

Given data:

Initial volume = 4.00 L

Initial pressure = 1.20 atm

Final volume = 2500  mL  

Final pressure = ?

Solution:

First of all we will convert the volume into litter.

2500 mL × 1L/1000 mL = 2.5 L

P₁V₁ =  P₂V₂

P₂ = P₁V₁ /V₂

1.20 atm×4.00 L = P₂ ×2.5 L

P₂ = 1.20 atm×4.00 L/ 2.5 L

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Answer:

 

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Explanation:

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