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Sunny_sXe [5.5K]
3 years ago
14

Describe and explain how the fractionating column is used to separate the mixture of hydrocarbon in crude oil ~ thank you

Chemistry
1 answer:
wolverine [178]3 years ago
5 0

Answer:

heated crude oil enters a tall fractionating column , which is hot at the bottom and gets cooler towards the top. vapours from the oil rise through the column. vapours condense when they become cool enough. liquids are led out of the column at different heights.

Explanation:

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A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi
Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of Na = 5.0 g

Mass of Cl_2 = 10.0 g

Molar mass of Na = 23 g/mol

Molar mass of Cl_2 = 71 g/mol

First we have to calculate the moles of Na and Cl_2.

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol

and,

\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}

\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

2Na+Cl_2\rightarrow 2NaCl

From the balanced reaction we conclude that

As, 2 mole of Na react with 1 mole of Cl_2

So, 0.217 moles of Na react with \frac{0.217}{2}=0.108 moles of Cl_2

From this we conclude that, Cl_2 is an excess reagent because the given moles are greater than the required moles and Na is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 0.217 mole of HCl react to give 0.217 mole of NaCl

Now we have to calculate the mass of NaCl

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

Molar mass of NaCl = 58.5 g/mole

\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g

Now we have to calculate the percent yield of this reaction.

Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = \frac{5.24g}{12.7g}\times 100

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0
4 years ago
Pls help!!<br>plz answer correctly!<br>will give the brainliest!<br>Urgent!!​
blagie [28]

a)

A: Copper

B: CuO

C: \mathrm{CuSO_4}

D: $\mathrm{CuCO_3}$

E: $\mathrm{CO_2}$

F: $\mathrm{Cu(NO_3)_2}$

b)

$\mathrm{CuO+ H_2SO_4}\rightarrow \mathrm{CuSO_4 + H_2O}$

c)

$\mathrm{CuCO_3+ 2HNO_3}\rightarrow \mathrm{Cu(NO_3)_2+ CO_2+ H_2O}$

7 0
3 years ago
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Determine a massa de um átomo de carbono12 em gramas
lapo4ka [179]
I believe it would be 1.660539040 × 10−24 gram.
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3 years ago
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Someone please help me with this ???
Verizon [17]

Answer:

c

Explanation:

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Which of these statements is false?
AVprozaik [17]

Answer:

The answer is b

Explanation:

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