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Tutor profile: Rosa Z.

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Rosa Z.
Chemical Engineering PhD Student
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Questions

Subject: Spanish

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Question:

El Jardín Cuelgan racimos de odorables pomas, negras uvas en gajos tentadores, fingiendo los alegres surtidores un murmullo de besos y de bromas. Dormitan en las ramas las palomas los buches esponjando arrulladores, y el capitoso aliento de las flores unge el follaje y el parral de aromas. Un sol ardiente esparce sus madejas de luz, sobre el jardín; y las abejas un vals preludian, áspero y sonoro. Bailan las mariposas deslumbrantes, y picotean pájaros brillantes unas naranjas que parecen de oro. - Juan Ramón Molina ¿Qué se puede concluir sobre la opinión del autor acerca del olor en el jardín? A. El autor piensa que los olores son desagradables. B. El autor piensa que los olores son muy refrescantes. C. El autor piensa que los olores son intensos. D. No podemos hacer ninguna conclusión acerca del olor en el jardín.

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Rosa Z.
Answer:

En el tercer y cuarto verso de la segunda estrofa el autor describe el olor de las flores como capitoso o caprichoso, e indica que los aromas también se han adherido a otras plantas en el jardín. Molina no incluye detalles sobre los aromas que indiquen si le parecen agradables o no. Sin embargo, las palabras que utiliza para describirlos demuestran que los aromas probablemente son bastante fuertes ya que sienten por todo el jardín. Por esta razón, la respuesta C seria correcta. No podemos elegir la respuesta B ya que no es posible saber si Molina piensa que los aromas son refrescantes basado en su descripción y no es correcto elegir la respuesta D porque si tenemos suficiente información para hace una conclusión. In the last two lines of the second stanza of this poem, the author describes the smell of the flowers as stubborn and indicates that it has spread over all the plants in the garden. Based on this description it would be correct to assume that Molina would consider the flower smell to be "intenso" or intense. He doesn't use adjectives that would suggest he finds these aromas unpleasant so we cannot choose option A. Similarly, we can't choose option B because his description does not give hints of the smell being refreshing. Option D would not be correct because his description does allow us to conclude that C is correct, meaning that D is incorrect.

Subject: Chemistry

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Question:

Calculate the change in standard gibbs free energy at 25$$^\circ$$C for the water-gas shift reaction: $$CO + H_2O \rightleftharpoons{} CO_2 + H_2$$ Given at 25$$^\circ$$C: $$\Delta H_{rxn}^\circ = -9.84 kcal$$

Inactive
Rosa Z.
Answer:

The change in gibbs free is given by: $$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$ We are given the change in enthalpy so we need to calculate the change in entropy. To do this, we can look up the entropy of the species involved in the reaction in online tables and find: $$S^\circ _{CO} = 197.6 \frac{J}{mol K} \hspace{1cm} S^\circ _{H_2O_{(g)}} = 188.7 \frac{J}{mol K} \hspace{1cm} S^\circ _{CO_{2}} = 213.6 \frac{J}{mol K}$$ $$S^\circ _{H_2} = 130.6 \frac{J}{mol K} $$ The change in entropy for a reaction is given by: $$\Delta S^\circ = \sum nS^\circ (products) - \sum nS^\circ (reactants)$$ Where n is the stoichiometric coefficient corresponding to the species in this reaction. Using this formula we get: $$\Delta S^\circ = [(1)(213.6) + (1)(130.6)] - [(1)(197.6) + (1)(188.7)] = -42.1 \frac{J}{K} $$ Now, to find T$$\Delta S ^\circ$$ by multiplying by the temperature, we first need to convert from C$$^\circ$$ to K. $$25 + 273.15 = 298.15 K$$ $$T\Delta S ^\circ = (298.15)(-42.1) = -12,552.1 J = -12.55 kJ$$ We now convert the enthalpy of reaction given from kcal to kJ to get: $$\Delta H_{rxn}^\circ = -9.84 kcal = -42.26 kJ$$ Now we substitute into the change of free energy equation: $$\Delta G^\circ = -42.26 - (-12.55) = -29.7 kJ $$

Subject: Calculus

TutorMe
Question:

Integrate the following: $$\int xe^{-x}dx$$

Inactive
Rosa Z.
Answer:

To integrate this expression we can use integration by parts. Quick reminder of the formula for integration by parts: $$\int u dv = uv - \int v du$$ We will assign u and v' to be: $$u = x \hspace{1cm} dv = e^{-x}$$ To evaluate the indefinite integral we determine du and v: $$du = 1dx \hspace{1cm} v = -e^{-x}$$ By substituting these into the integration by parts formula and get: $$\int xe^{-x} dx = -xe^{-x} - \int (-e^{-x})dx = -xe^{-x}+\int e^{-x}dx$$ $$= -xe^{-x} - e^{-x} + C$$ Where C is an undefined constant.

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