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Jasper S.
Dean's List Student at The University of Chicago
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Spanish
TutorMe
Question:

Translate: I want to go to the market. When I was five I wanted to go to the beach. I will visit you Tuesday.

Jasper S.
Answer:

I want to go to the market. This is a present tense desire so we can simply use the present tense. -Yo quiero ir al mercado. When I was five I wanted to go to the beach. This is a past tense desire so we have to use the imperfect tense. -cuando yo tenía cinco años que quería ir a la playa. I will visit you Tuesday. This is a future tense sentence so we can use the simple future but we also have a direct object we can incorperated as a direct object pronoun as well. -Yo te visitaré Martes.

Chemistry
TutorMe
Question:

What is the maximum mass of Iron(iii) oxide [Fe2O3] which can be produced when 6 moles of iron is exposed to excess oxygen.

Jasper S.
Answer:

Before we begin our stoichiometry it is important to write the balanced equation for the reaction. -4Fe + 3O2 -> 2Fe2O3 The limiting reagent in the problem is Fe so we begin our stoichiometry by calculating the maximum number of moles of Fe2O3 that can be produced with from 6 moles of Fe. -According to our balanced equation, for every 2 moles of Fe we can produce 1 mole of Fe2O3. We have 6 moles of Fe so we can produce a maximum of 3 moles of Fe2O3. Now that we know how many moles of Fe2O3 that can be produced all that is left is to calculated the mass assosciated with that number of moles. -The molar mass of Fe is 55.845g and the molar mass of O is 15.999g therefor the molar mass of Fe2O3 is 55.845(2)g + 15.999(3)g = 159.687g Finally to calculate the mass of Fe2O3 that is formed from the reaction of 6 moles of Fe in excess oxygen we multiply the molar mass of Fe2O3 by the maximum number of moles that can be created. -(159.687g) x 3 = 479.061g

Calculus
TutorMe
Question:

Given: f(x)=3x+3 Find the area beneath the graph between x=1.5 and x=2.5

Jasper S.
Answer:

To solve this problem you must begin by integrating the formula given. The integration of 3x+3 is (3/2)x^2+3x. Next the values can be substituted into the integrated formula. For the first value (3/2)(2.5^2) + 3(2.5) = (3/2)(6.25) + 7.5 = 9.375 + 7.5 = 16.875 For the second value (3/2)(1.5^2) + 3(1.5) = (3/2)(2.25) + 4.5 = 3.375 + 4.5 = 7.875 Now we subtract the value of the higher bound from that of the lower bound to receive the final answer for the area under the graph. 16.875 - 7.875 = 9

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