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ANTONII [103]
3 years ago
14

What's a petroleum industriously​

Chemistry
2 answers:
scoray [572]3 years ago
8 0

Answer:

Given to industry; acting or working with diligence; sedulous: as, a person industrious in business.

Marked by industry; done with or characterized by diligence; busily pursued, performed, or employed: as, an industrious life; industrious researches.

Expert; clever; shrewd.

Hope this helped

Explanation:

Citrus2011 [14]3 years ago
6 0

Answer: Petroleum is a naturally occurring liquid found beneath the earth’s surface that can be refined into fuel. Petroleum is a fossil fuel, meaning that it has been created by the decomposition of organic matter over millions of years. Petroleum is formed when large quantities of dead organisms–primarily zooplankton and algae–underneath sedimentary rock are subjected to intense heat and pressure.

Explanation:

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Please help with this question
tamaranim1 [39]
1.806x10^24
Written equation form(always start the equation off with what you know based off of the question!):

3mol(CCl4)•6.022x10^23/1mol = 1.806x10^24

Good luck!
3 0
3 years ago
Why is monatomic compound nonsense?
Tamiku [17]
<span>Because mona means 1 atom, and a compound is made up of 2 or more atoms.</span>
3 0
3 years ago
Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
3 years ago
The 3 components of air
jok3333 [9.3K]

Nitrogen (around 78%), Oxygen (around 21%), and Argon (around 1%).

Hope this helps :)

8 0
2 years ago
Read 2 more answers
What are some problems chemist must consider when developing new technologies?
Ratling [72]
The study of chemistry helps us understand the nature of the world around us. Chemistry is always developing to keep up with any phenomenon that appears in nature.
Therefore, scientists and chemists are always developing new technologies. However, chemists must very careful when developing these new technologies. They should consider any bad chemical reactions that might occur and also chemicals that harmful to either the individuals or the environment.
3 0
3 years ago
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