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3 years ago

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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Explain why calcium (Ca) has a greater atomic radius than magnesium (Mg).

g100num [7]

Answer:

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2 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

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2 years ago

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2 years ago

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3 years ago

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2 years ago

What were the limitations of Dobereiner's classifcation?​

What were the limitations of Dobereiner's classifcation?