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irina1246 [14]
1 year ago
14

What visual criterion is to be used to assess whether the crude reaction product is dry prior to semimicro distillation

Chemistry
1 answer:
Olenka [21]1 year ago
5 0

The visual criterion which has to be assessed  is to ensure it's not oily or sticky.

<h3>What is Distillation?</h3>

This is a separation technique that involves the use selective boiling and condensation to separate a liquid mixture into different components.

In the case of Crude reaction product , it is best to ensure that it is not oily or sticky before semimicro distillation is done.

Read more about Distillation here brainly.com/question/552187

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The most important commercial ore of chromium is chromite (FeCr2O4). One of the steps in the process used to extract chromium fr
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Answer:

1 mole FeCr2/ 1 mole FeCr2O4

Explanation:

We have to write down the equation of the reaction before we can answer the question;

2C(s) + FeCr2O4(s) -------> FeCr2(s) + 2CO2(g)

By inspection of this reaction equation, we can clearly see that the mole ratio of the reactants and products is 2:1:1:2.

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The mole ration required to convert chromites to ferrochrome is;

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2 years ago
what is the percent yield of titanium (II) oxide if 20.0 grams of titanium (II) sulfide is reacted with water? The actual yield
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Answer : The percent yield of titanium (II) oxide is, 142.5 % and the impurities could have caused the percent yield to be so high.

Explanation : Given,

Mass of titanium(II) sulfide = 20.0 g

Molar mass of titanium(II) sulfide = 79.9 g/mole

Molar mass of titanium(II) oxide = 63.9 g/mole

First we have to calculate the moles of titanium(II) sulfide.

\text{ Moles of titanium(II) sulfide}=\frac{\text{ Mass of titanium(II) sulfide}}{\text{ Molar mass of titanium(II) sulfide}}=\frac{20.0g}{79.9g/mole}=0.2503moles

Now we have to calculate the moles of titanium(II) oxide.

The balanced chemical reaction is,

TiS+H_2O\rightarrow TiO+H_2S

From the reaction, we conclude that

As, 1 mole of titanium(II) sulfide react to give 1 mole of titanium(II) oxide

So, 0.2503 mole of titanium(II) sulfide react to give 0.2503 mole of titanium(II) oxide

Now we have to calculate the mass of titanium(II) oxide.

\text{ Mass of titanium(II) oxide}=\text{ Moles of titanium(II) oxide}\times \text{ Molar mass of titanium(II) oxide}

\text{ Mass of titanium(II) oxide}=(0.2503moles)\times (63.9g/mole)=15.99g

To calculate the percentage yield of titanium (II) oxide, we use the equation:

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

Experimental yield of titanium (II) oxide = 22.8 g

Theoretical yield of titanium (II) oxide = 15.99 g

Putting values in above equation, we get:

\%\text{ yield of titanium (II) oxide}=\frac{22.8g}{15.99g}\times 100\\\\\% \text{yield of titanium (II) oxide}=142.5\%

Hence, the percent yield of titanium (II) oxide is, 142.5 %

If the percent yields is greater than 100% that means the product of the reaction contains impurities which cause its mass to be greater than it actually.

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