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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.”

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HNO3 and H2CO3 are examples of ?

Basile [38]

A) acids because they start with h

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3 years ago

What are the signs of the enthalpy change (ΔH°) and the entropy change (ΔS°) for the condensation of CS2(g)?

VARVARA [1.3K]

Answer:

∆H is negative

∆S is negative

Explanation:

The condensation of CS2 implies a phase change from gaseous state to liquid state. The energy of the gaseous particles is greater than that of the liquid particles hence energy is given out when a substance changes from gaseous state to liquid state hence the process is exothermic and ∆H is negative.

Changing from gaseous state to liquid states leads to a decrease in entropy hence ∆S is negative. Liquid particles are more orderly than particles of a gas.

5 0

4 years ago

What is the difference between covalent bonding and electrovalent bonding

Rudiy27

Answer:

1 ) Electrovalent compounds are formed by the complete transfer of electrons while covalent compounds are formed by sharing of electrons between 2 atoms.

2) Electrovalent compounds are more soluble in polar solvents like water while covalent compounds are more soluble in non-polar solvents like methane.

5 0

3 years ago

An aqueous solution 10 g of an optically pure substance diluted to 500ml with water and placed in a polarimeter tube 20 cm long.

Gennadij [26K]

Explanation:

Formula to calculate specific rotation is as follows.

Specific rotation ( $[\alpha]$ ) = $\frac{\alpha}{c} \times l$

where, $\alpha$ = observed rotation

c = concentration in g/ml

l = path length in dm

It is given that,

$\alpha = -6.16^{o}$

c = $\frac{10 g}{500 ml}$ = 0.02 g/ml

l = 20 cm = 2 dm (as 1 dm = 10 cm)

Therefore, calculate the specific rotation as follows.

Specific rotation ( $[\alpha]$ ) = $\frac{\alpha}{c} \times l$

= $\frac{-6.16^{o}}{0.02 g/ml} \times 2 dm$

= $-616^{o}$

Thus, we can conclude that the specific rotation of this compound is $-616^{o}$ .

7 0

3 years ago

Learn how changes in binding free energy affect binding and the ratio of unbound and bound molecules.

Delvig [45]

C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

E)[D']_T = 1.495* 10 ^-7 M

F)[D'] / [ED'] = 0.0579

Explanation:

E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K

Dissociation constant of K_D = 1.3 μM (1.3 *10 ^-6 M)

E + D ⇄ ED → K_a = [ED] / [D][E] (association constant)

ED ⇄ E + D → K_D = [E][D] / [ED] (dissociation constant)

[E] =2.5*10^-7 mol/L

K_D = 1.3* 10^-6 M

K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]

= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7

= 13/25 * 10

=130/25 = 5.20

[D]/[ED] = 5.20

ΔG =RTln Kd

ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6

ΔG_2 592.454 * [ln 1.3 +ln 10^-6]

ΔG_1 = 592.424 [0.2623 - 13.8155]

ΔG_2 = -592.424 * 13.553

ΔG_1 = -8184.633 cal/ mol

ΔG_1 = -8184.633 * 4.18 J/mol = -34244.508 J?mol

ΔG_1 = -34.245 KL/mol

so, ΔG_2 = ΔG_1 - 10.5 KJ/mol

ΔG_2 = -34.245 - 10.5

ΔG_2 = -44.745KJ / mol

ΔG_2 =RT ln K_D

-44.745 *10^3

=8.314 *298.15 lnK_D

lnK_D' = - 44745 / 2478.81 g

ln K_D' = -18.051

K_D' = -18.051

K_D' = e^-18.051

[D]/[ED] = 5.20

[E] = 2.5* 10 ^-7 mol/ L = a

K_D' = [E][D] / [ED'] E +D' → ED'

K_D' = a/2(x-(a/2) / (a/2)

KD' = x - a/2

=2.447 *10^-8 = (2.5/2) * 10^-7

x=2.447 * 10^-8 + 1.25 * 10^-7

x = 2.447 *10^-8 + 1.25 * 10 ^-7

x= 10^-7 [1.25 + 0.2447]

x = 1.4947 * 10^-7

[D']_T = 1.495* 10 ^-7 M

K_D' = [E][D'] / [ED']

[D'] / [ED'] = KD' / [E]

[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7

[D'] / [ED'] = 0.5788 * 10^-1

[D'] / [ED'] = 0.0579

5 0

3 years ago