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# Tutor profile: Simon B.

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Simon B.
Dental Student
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## Questions

### Subject:Physics (Thermodynamics)

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Question:

What is ∆G and how can we predict the spontaneity of a reaction using entropy and enthalpy?

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Simon B.

∆G is a measure of free-energy and if the calculated value is negative, we can conclude that the reaction is spontaneous in the forward direction. The equation for Gibbs free-energy is ∆G = ∆H - T∆S. This equation is very powerful because we can predict the spontaneity of a reaction if we know our change in enthalpy and entropy. For instance, if there is a very large negative change in enthalpy, we can predict that the free energy will also be negative and result in a spontaneous reaction. Similar information can be obtained when working with temperature or the change in entropy.

### Subject:Organic Chemistry

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Question:

In an addition reaction where we add HBr to an alkene, why does the Br end up on the more substituted carbon as opposed to the less substituted carbon?

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Simon B.

The answer is due to the stability of the intermediate structure. When the electrons on the double bond "grab the hydrogen from HBr, it forms a carbocation. Since more substituted carbocations are more stable, the H will be added in a way that forms a more favorable carbocation (tertiary > secondary > primary). Once the carbocation is formed, the electrons from Br- will attack the carbocation and this is why Br ends up on the more substituted carbon.

### Subject:Chemistry

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Question:

If you take X grams of compound A and mix it with Y grams of compound B, how much of the product, C, will be generated from the reaction?

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Simon B.

In order to solve this problem, you need to first ask yourself, "What is the limiting reagent?" To do this, start by converting the grams of compound A and compound B into moles. To do this, you must look up the molar mass of each compound and divide the initial amount (in grams) by the molar mass (grams/mole). Next, use the reaction coefficients to relate compound A with compound C and compound B with compound C. Since you are solving for compound C, you will put the moles of compound C in the numerator and Compound A or B in the denominator. Finally, multiply each by the molar mass of compound C. This will give you two answers, the smaller one being the correct one because it factors in the limiting reagent.

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