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3 years ago

What are the sources of petrochemicals

Chemistry

2 answers:

ollegr [7]3 years ago

6 0

Answer:

The sources of petrochemical are: fossil fuel such as coal and natural gases or from renewable sources such corn, sugar, cane and other types of biomass .

Explanation:

If this help you then mark me as a brainliest

Sindrei [870]3 years ago

4 0

Answer:

Petrochemicals are chemical products derived from petroleum, although many of the same chemical compounds are also obtained from other fossil fuels such as coal and natural gas or from renewable sources such as corn, sugar cane, and other types of biomass.

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Which of the following equations has the correct products and is balanced correctly for a reaction between Na3PO4 and KOH?

ASHA 777 [7]

The answer is: A) Na3PO4 + 3KOH → 3NaOH + K3PO4, because K retains the same charge throughout the reaction.

This chemical reaction is double displacement reaction - cations (K⁺ and Na⁺) and anions (PO₄³⁻⁻ and OH⁻) of the two reactants switch places and form two new compounds.

Na₃PO₄ is sodium phosphate.

KOH is potassium hydroxide.

NaOH is sodium hydroxide.

K₃PO₄ is potassium phosphate.

According to the mass conservation law, there are same number of atoms on both side of balanced chemical reaction.

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4 years ago

Provide the quantum numbers for each electron in the element gadolinium

Basile [38]

Atomic # 64
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3 years ago

Why could a diet low in carbohydrates be dangerous?

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4 years ago

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3 years ago

The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3⋅xH2O The number of kilowatt-hours o

vesna_86 [32]

Explanation:

According to Faraday's law, the amount of a substance deposited or liberated in electrolysis process is proportional to the quantity of electric charge passed and to the equivalent weight of the substance.

Formula to calculate the mass of substance liberated according to Faraday's law is as follows.

m = $(\frac{Q}{F})(\frac{M}{Z})$

where, m = mass of substance liberated at electrode

Q = electric charge passing through the substance

F = Faraday constant = 96,487 C $mol^{-1}$

M = molar mass of the substance

Z = valency number of ions of the substance

Since, it is given that mass is 3 kg or 3000 g (as 1 kg = 1000 g), molar mass of Al is 27, Z is 3.

Therefore, putting the values in the above formula as follows.

m = $(\frac{Q}{F})(\frac{M}{Z})$

3000 g = $(\frac{Q}{96,487 C mol^{-1}})(\frac{27}{3})$

Q = 32162333.33 C

As it is given that V = 4.50 Volt. Also, it is known that

Energy = $V \times Q$

Therefore, calculate the energy as follows.

Energy = $V \times Q$

= $4.50 V \times 32162333.33 C$

= 144730500 J

As it is known that $3.6 \times 10^{6}$ J = 1 KW Hr

So, convert 144730500 J into KW Hr as follows.

$\frac{144730500 J \times 1 KW Hr}{3600000 J}$

= 40.202 KW Hr

Thus, we can conclude that the number of kilowatt-hours of electricity required to produce 3.00 kg of aluminum from electrolysis of compounds from bauxite is 40.202 KW Hr when the applied emf is 4.50V.

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3 years ago