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Lady_Fox [76]
3 years ago
9

Which observation would most likely indicate that an endothermic reaction has occurred? O A. a color change O B. the production

of gas bubbles O C. an increase in temperature OD. a decrease in temperature​
Chemistry
1 answer:
Marina CMI [18]3 years ago
7 0

Answer:

that's allow points

........

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Where do the lemons go after harvested? What gets done with lemons?
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Lemon juice,lemonade

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2 years ago
Using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction free energy of the following chemical
julia-pushkina [17]

Answer:

43.0 kJ

Explanation:

The free energy (ΔG) measures the total energy that is presented in a thermodynamic system that is available to produce useful work, especially at thermal machines. In a reaction, the value of the variation of it indicates if the process is spontaneous or nonspontaneous because the free energy intends to decrease, so, if ΔG < 0, the reaction is spontaneous.

The standard value is measured at 25°C, 298 K, and the value of free energy varies with the temperature. It can be calculated by the standard-free energy of formation (G°f), and will be:

ΔG = ∑n*G°f products - ∑n*G°f reactants, where n is the coefficient of the substance in the balanced reaction.

By the balanced reaction given:

2NOCl(g) --> 2NO(g) + Cl2(g)

At ALEKS Data tab:

G°f, NOCl(g) = 66.1 kJ/mol

G°f, NO(g) = 87.6 kJ/mol

G°f, Cl2(g) = 0 kJ/mol

ΔG = 2*87.6 - 2*66.1

ΔG = 43.0 kJ

6 0
3 years ago
The activation energy for the isomerization ol cyclopropane to propene is 274 kJ/mol. By what factor does the rate of this react
Bond [772]

Answer:

The rate of the reaction increased by a factor of 1012.32

Explanation:

Applying Arrhenius equation

ln(k₂/k₁) = Ea/R(1/T₁ - 1/T₂)

where;

k₂/k₁ is the ratio of the rates which is the factor

Ea is the activation energy = 274 kJ/mol.

T₁ is the initial temperature = 231⁰C = 504 k

T₂ is the final temperature = 293⁰C = 566 k

R is gas constant = 8.314 J/Kmol

Substituting this values into the equation above;

ln(k₂/k₁) = 274000/8.314(1/504 - 1/566)

ln(k₂/k₁) = 32956.4589 (0.00198-0.00177)

ln(k₂/k₁)  = 6.92

k₂/k₁ = exp(6.92)

k₂/k₁ = 1012.32

The rate of the reaction increased by 1012.32

3 0
3 years ago
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