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Xelga [282]
3 years ago
12

Cite as principais características das tentativas de organização da Tabela periódica-Tríades de Dobereiner, o Parafuso Telúrico

de Chancourtois e a Lei das Oitavas de Newlands.
Chemistry
1 answer:
sdas [7]3 years ago
6 0

Answer:

Todos os três cientistas classificaram os elementos de acordo com o peso atômico.

Explicação:

As características da Tabela Periódica da Tríade Dobereiner são as seguintes:

1. é uma das classificações mais antigas de elementos.

2. nesta classificação, o peso atômico de um átomo é igual à média dos outros dois pesos atômicos.

3. é baseado no peso atômico.

Características do parafuso telúrico de Chancourtois

1. É a primeira classificação baseada no peso atômico.

2. foi a primeira representação gráfica dos elementos da tabela periódica.

Características da Lei das Oitavas de Newlands

1. também organizou os elementos de acordo com o peso atômico.

2. Em uma linha há apenas sete elementos e o oitavo vem abaixo do primeiro elemento devido às mesmas propriedades físicas e químicas.

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Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o
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<u>Answer:</u> The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

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To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of CrO_2 = 480.1 g

Molar mass of CrO_2 = 84 g/mol

Putting values in equation 1, we get:

\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol

For the given chemical equation:

4CrO_2\rightarrow 2Cr_2O_3+O_2

By Stoichiometry of the reaction:

4 moles of CrO_2 produces 2 moles of chromium (III) oxide

So, 5.72 moles of CrO_2 will produce = \frac{2}{4}\times 5.72=2.86mol of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g

To calculate the percentage yield of chromium (III) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

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