For any spontaneous process, universe entropy intensifies is known as the second law of thermodynamics.
<h3>What is entropy?</h3>
Entropy is defined as the degree of randomness or disorderliness of a system.
The entropy of a system generally increases for any spontaneous process.
This is according to the second law of thermodynamics.
In conclusion, the entropy of a system is the a measure of randomness of the system.
Learn more about entropy at: brainly.com/question/21578229
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Answer:
Explanation:
The metric system is a system of measurement that uses the meter, liter, and gram as base units of length (distance), capacity (volume), and weight (mass) respectively.
To measure smaller or larger quantities, we use units derived from the metric units
metric-system
The given figure shows the arrangement of the metric units, which are smaller or bigger than the base unit.
The units to the right of the base unit are smaller than the base unit. As we move to the right, each unit is 10 times smaller or one-tenth of the unit to its left. So, a ‘deci’ means one-tenth of the base unit, ‘centi’ is one-tenth of ‘deci’ or one-hundredth of the base unit and ‘milli’ is one-tenth of ‘centi’ or one-thousandth of the base unit.
The units to the left of the base unit are bigger than the base unit. As we move to the left, each unit is 10 times greater than the unit to its right. So, a ‘deca’ means ten times of the base unit, ‘hecto’ is ten times of ‘deca’ or hundred times of the base unit and ‘killo’ is ten times of ‘hecto’ or thousand times of the base unit.
Answer:
ΔH°r = -1562 kJ
Explanation:
Let's consider the following combustion.
C₂H₆(g) + 7/2 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(l)
We can calculate the standard heat of reaction (ΔH°r) using the following expression:
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
where,
ni are the moles of reactants and products
ΔH°f(i) are the standard heats of formation of reactants and products
The standard heat of formation of simple substances in their most stable state is zero. That means that ΔH°f(O₂(g)) = 0
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
ΔH°r = [2 mol × ΔH°f(CO₂) + 3 mol × ΔH°f(H₂O)] - [1 mol × ΔH°f(C₂H₆) + 7/2 mol × ΔH°f(O₂)]
ΔH°r = [2 mol × (-394.0 kJ/mol) + 3 mol × (-286.0 kJ/mol)] - [1 mol × (-84.00 kJ/mol) + 7/2 mol × 0]
ΔH°r = -1562 kJ
