No, the properties of a substance are not affected by the amount of a substance.
Answer:
different reactants
Explanation:
because if you think about it what's really thick and it takes a lot longer to burn a piece of paper is super fit material which can just tear real easily just by ripping it with your fingers so if you think about it how fast it burns it's not how big the molecules are because paper has smaller molecules than wood would so the paper burns faster in differently so it's probably the reactant it's made of
Answer: Free energy combined enthalpy and entropy into a single value. Gibbs’s free energy is the energy associated with a chemical reaction that can do useful work. It equals the enthalpy minus the product of temperature and entropy of the system.
Explanation:
Answer:
C
Explanation:
Temperature is directly related to kinetic energy (KE). As we raise temperature, we are raising KE, as well. Particles with more KE move more quickly and with more force.
This means that these particles are more likely to collide with each other and react to allow the chemical reaction to follow through. In turn, if the chemical reaction is more likely to go to completion, the reaction rate increases, eliminating A and B.
The concentration of the solute is not affected by the temperature; in other words, temperature will not increase or decrease the amount of solute in the solution, so eliminate D.
Thus the answer is C.
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Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
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