All categories

3 years ago

What is the main "impurity" collected with the oil?

Chemistry

1 answer:

NikAS [45]3 years ago

7 0

Answer:

Sulfur is considered an “impurity” in petroleum. Sulfur in crude oil can corrode metal in the refining process and contribute to air pollution. Petroleum with more than 0.5% sulfur is called “sour,” while petroleum with less than 0.5% sulfur is “sweet.”

You might be interested in

Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2

PIT_PIT [208]

<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ$

Hence, the standard heat for the given reaction is -138.82 kJ

3 0

3 years ago

What is the mass of CO2 lost at 20 min from the limestone sample?

harkovskaia [24]

Answer:

The mass, CO2 and CO3 from the limestone sample is discussed below in details.

Explanation:

(A) mass loss of sample of limestone after 20 min

= 0.8437g-0.5979g = 0.2458 g

From the given reaction of limestone, 2 mol of the sample gives 2 moles of CO 2.

Therefore

184.4 g ( molar mass of limestone) gives2× 44 g of carbon dioxide.

1 g of sample gives 88/184.4 g of carbon dioxide

Hence 0.2458 g sample gives

= 88/184.4 × 0.2458 g = 0.117 g carbon dioxide

(B) mole of CO 2 lost = weight/ molar mass

= 0.117 g / 44 g/mol =0.0027 mole

(C). 1 mol of limestone contain 2 mol of carbonate ion

From the reaction we know that carbonate ion of limestone is converted into carbondioxide

Hence lost carbonate ion = 0.2458 g

(D) we know that

1 mol limestone contain 1mol CaCO 3

Hence in sample present CaCO 3

= 1mole / 184.4 g × 0.8437 g= 0.00458 mol CaCO3

8 0

3 years ago

Write L values for the following energy levels. What orbital shapes exist in each of

den301095 [7]

Answer:

n l m

��

1 0 0 1s 1 2 2

��

2 0 0 2s 1 2

2 1 1,0,-1 2p 3 6 8

��

3 0 0 3s 1 2

3 1 1,0,-1 3p 3 6

3 2 2,1,0,-1,-2 3d 5 10 18

��

4 0 0 4s 1 2

4 1 1,0,-1 4p 3 6

4 2 2,1,0,-1,-2 4d 5 10

4 3 3,2,1,0,-1,-2,-3 4f 7 14 32

Explanation:

4 0

2 years ago

45.7 grams of calcium chloride reacts with an excess of aluminum oxide. How many grams of aluminum chloride will be produced

damaskus [11]

Molar mass (CaCl2) = 40.1 +2*35.5 = 111.1 g/mol
Molar mass (AlCl3) = 27.0 +3*35.5= 133.5 g/ mol

3CaCl2+Al2O3 -------->3CaO +2AlCl3
mole from reaction 3 mol 2 mol
mass from reaction 3mol* 111.1g/mol 2 mol*133.5g/mol
333.3 g 267.0 g
mass from problem 45.7 g x g

Proportion:
333.3 g CaCl2 ------- 267.0 g AlCl3
45.7 g CaCl2 -------- x g AlCl3

x=45.7*267.0/333.3= 36.6 g AlCl3

5 0

3 years ago

The process in which water changes from a liquid to a vapor is known as __________.

Basile [38]

Hello!
The answer is Evaporation!

When you add enough heat to a liquid, it boils turning into a gas. This is called Evaporation

7 0

3 years ago

Read 2 more answers