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2 years ago

Separating techniques - Chromatography

Chemistry

1 answer:

vodomira [7]2 years ago

7 0

The method that can be used to separate the mixture is chromatography.

<h3>What is chromatography?</h3>

"Chromatography" is obtained form a Greek word which literarily means, color writing. It is a method of separation which is common in separating a mixture of pigments.

To obtain the colors used, two solvents are mixed and the sample ink is dissolved in the solvents then spotted on a thin layer and put into a TLC chamber then the chromatogram is allowed to develop.

The various components of the pigment will appear on the chromatogram and can be identified using spectrophotometry. The Rf values of each component can also be used to identify it.'

Learn more about chromatography: brainly.com/question/26491567

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Name the alcohol that contains five carbons and has a hydroxyl (alcohol) group on the second carbon.

Musya8 [376]

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4 years ago

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3 years ago

2Al + 3MnSO4 ↔ Al2(SO4)3 + Mn

Stells [14]

In accordance with the Le Chatelier's Principle, when the concentration of the aluminum increases then the equilibrium would shift to the right. It will produce more products since there are more reactants available and also collision is more frequent.

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4 years ago

Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3

AveGali [126]

Answer : The percent yield of $CO_2$ is, 68.4 %

Solution : Given,

Mass of $CH_4$ = 0.16 g

Mass of $O_2$ = 0.84 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$ .

$\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 0.026 moles of $O_2$ react with $\frac{0.026}{2}=0.013$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $O_2$ react to give 1 mole of $CO_2$

So, 0.026 moles of $O_2$ react to give $\frac{0.026}{2}=0.013$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$

$\text{ Mass of }CO_2=(0.013moles)\times (44g/mole)=0.572g$

Theoretical yield of $CO_2$ = 0.572 g

Experimental yield of $CO_2$ = 0.391 g

Now we have to calculate the percent yield of $CO_2$

$\% \text{ yield of }CO_2=\frac{\text{ Experimental yield of }CO_2}{\text{ Theretical yield of }CO_2}\times 100$

$\% \text{ yield of }CO_2=\frac{0.391g}{0.572g}\times 100=68.4\%$

Therefore, the percent yield of $CO_2$ is, 68.4 %

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3 years ago

Electronegativity values increase from left to right across a period in the periodic table because there is an increase in the c

Mademuasel [1]

First, let us define Electronegativity. Electronegativity is "the ability of an atom to attract electrons." In addition, electronegativity increases in elements from left to right, while on the other hand, electronegativity decreases from top to bottom in an element group. It decreases because the atomic radius increases as we go downward an element in the group.

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4 years ago