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2 years ago

Q2. Name, constituent of petroleum (Petrochemicals) which is used for following purposes.

Chemistry

1 answer:

finlep [7]2 years ago

7 0

The constituents of petroleum that are used for the following purposes are as follows:

To make candles ----- Paraffin wax
A solvent for dry cleaning ----- Petrol
For surfacing roads ----- Bitumen
Jet engine fuel ----- Kerosene
For lubrication ----- Lubricating oil.

<h3>What are the constituents of petroleum?</h3>

The constituents of petroleum are LPG, bitumen, paraffin wax, lubricating oil, kerosene, diesel, etc. These compounds are a mixture of hydrocarbons.

Therefore, each constituent of petroleum that is used for the following purposes is mentioned above with proper names.

To learn more about constituents of petroleum, refer to the link:

brainly.com/question/17023299

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One of the intermediates in the synthesis of glycine from ammonia, carbon dioxide, and methane is aminoacetonitrile, C2H4N2. The

sleet_krkn [62]

Answer:

Mass of C₂H₄N₂ produced = 3.64 g

Explanation:

The balanced chemical equation for the reaction is given below:

3CH₄ (g) + 5CO₂ (g) + 8NH₃ (g) → 4C₂H₄N₂ (g) + 10H₂O (g)

From the equation, 3 moles of CH₄ reacts with 5 moles of CO₂ and 8 moles of NH₃ to produce 4 moles of C₂H₄N₂ and 10 moles of H₂O

Molar masses of the compounds are given below below:

CH₄ = 16 g/mol; CO₂ = 44 g/mol; NH3 = 17 g/mol; C₂H₄N₂ = 56 g/mol; H₂O g/mol

Comparing the mole ratios of the reacting masses;

CH₄ = 1.65/16 = 0.103

CO₂ = 13.5/44 = 0.307

NH₃ = 2.21/17 = 0.130

converting to whole number ratios by dividing with the smallest ratio

CH₄ = 0.103/0.103 = 1

CO₂ = 0.307/0.103 = 3

NH₃ = 0.130/0.103 = 1.3

Multiplying through with 5

CH₄ = 1 × 5 = 5

CO₂ = 3 × 5 = 15

NH₃ = 1.3 × 5 = 6.5

Therefore, the limiting reactant is NH₃

8 × 17 g (136 g) of NH₃ reacts to produce 4 × 56 g (224 g) of C₂H₄N₂

Therefore, 2.21 g of NH₃ will produce (2.21 × 224)/136 g of C₂H₄N₂ = 3.64 g of C₂H₄N₂

Mass of C₂H₄N₂ produced = 3.64 g

7 0

3 years ago

8.0 mol AgNO3 reacts with 5.0 mol Zn in

Yakvenalex [24]

Taking into account the reaction stoichiometry, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgNO₃ + Zn → 2 Ag + Zn(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

AgNO₃: 2 moles
Zn: 1 mole
Ag: 2 moles
Zn(NO₃)₂: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Zn reacts with 2 moles of AgNO₃, 5 moles of Zn reacts with how many moles of AgNO₃?

$amount of moles of AgNO_{3}= \frac{5 moles of Znx2 moles of AgNO_{3}}{1 mole of Zn}$

<u><em>amount of moles of AgNO₃= 10 moles </em></u>

But 10 moles of AgNO₃ are not available, 8 moles are available. Since you have less moles than you need to react with 5 moles of Zn, AgNO₃ will be the limiting reagent.

<h3>Moles of Ag formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 2 moles of AgNO₃ form 2 moles of Ag, 8 moles of AgNO₃ form how many moles of Ag?

$amount of moles of Ag=\frac{8 moles of AgNO_{3}x2 moles of Ag }{2 moles of AgNO_{3}}$

<u><em>amount of moles of Ag= 8 moles</em></u>

Then, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

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4 0

2 years ago

If the actual yield of a reaction is 37.6 g while the theoretical yield is 112.8 g what is the percent yield

Zigmanuir [339]

<h2>Hello!</h2>

The answer is:

The percent yield of the reaction is 32.45%

To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.

We are given that:

$ActualYield=37.6g\\TheoreticalYield=112.8g$

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.

So, calculating we have:

$PercentYield=\frac{ActualYield}{TheoreticalYield}*100\\\\PercentYield=\frac{37.6g}{112.8g}*100=0.3245*100=32.45(percent)$

Hence, we have that the percent yield of the reaction is 32.45%.

Have a nice day!

8 0

3 years ago

Read 2 more answers

How many atoms are in 8.28 moles of aluminum?

Lilit [14]

Answer:

49.86 × 10²³ atoms of Al

Explanation:

Given data:

Number of moles of Al = 8.28 mol

Number of atoms = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For 8.28 moles of Al:

1 mole = 6.022 × 10²³ atoms of Al

8.28 mol×6.022 × 10²³ atoms / 1mol

49.86 × 10²³ atoms of Al

4 0

3 years ago

The surface temperature on Venus may approach 739 K. What is this temperature in degrees Celsius?

Maslowich

739 degrees K equals 465.85 degrees Celcius.

7 0

3 years ago