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just olya [345]
3 years ago
10

1. What is the difference between a synthetic resource and a Natural Resource? This is technology.and I need help pls help!!!!!!

!!!!!!!
Chemistry
1 answer:
Elodia [21]3 years ago
4 0

Answer:

Natural materials are those that are found in nature and have not been made by humans. By comparison, synthetic materials are man-made and cannot be found in nature. Synthetic products are usually created in laboratories by mixing different chemicals, or prepared compounds and substances made in a laboratory.

Explanation:

Natural materials are those that are found in nature and have not been made by humans. By comparison, synthetic materials are man-made and cannot be found in nature. Synthetic products are usually created in laboratories by mixing different chemicals, or prepared compounds and substances made in a laboratory

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The light bends

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When it passes through two different densities it changes directions, causing a bend

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Which of the following are forms that chemical hazards can be in?
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How many significant figures in 11 soccer players
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Counting gives an exact number and exact numbers have infinite sig figs.
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What is the equilibrium equation for the dissociation of formic acid in water? hcooh (aq) + h2o (l) ⇌ h3o+ (aq) + hcoo- (aq)?
Pani-rosa [81]
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7 0
3 years ago
Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)
Vsevolod [243]

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

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

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

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

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K

ΔG° = 131.5 kJ

6 0
3 years ago
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