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Jeremiah B.
Chemical Engineer
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Chemistry
TutorMe
Question:

Hi! Would you help me with understanding the difference between an Sn1 and Sn2 reaction?

Jeremiah B.
Answer:

Absolutely! An Sn1 reaction occurs in 2 steps despite its name. The first step involves a leaving group leaving, typically a halogen like bromine. This forms a carbocation as well as a flat molecule. Be careful, if there is a nearby more stable carbon, a methyl or hydrogen may migrate to place the cation charge on a more stable carbon. A nucleophile (a group that attacks a positive charge) can attack that carbocation from either side and form a bond with the carbon which causes a scramble of the chiralty (mix of R and S). An Sn2 reaction occurs in 1 step. A nucleophile attacks the same carbon as the leaving group departs but from the opposite side. The leaving group leaves while the nucleophile bonds. This causes the chiralty to always invert ( R becomes S or S becomes R). No carbocation is ever formed. Sn1 prefers tertiary carbons since a carbocation is formed. Sn2 prefers primary carbon since a carbocation isn't formed and primary carbons aren't hindered by other groups. Hope this helps ya!

Calculus
TutorMe
Question:

Hello, could you explain to me how I could setup a triple integral to find the volume of solid bounded below by the sphere ρ= 4 cos(Φ) and above by the cone Φ = π/4?

Jeremiah B.
Answer:

No problem! With the given problem, I would recommend using spherical coordinates. Now, the general form of a triple integral using spherical coordinates is ∫∫∫ (ρ^2) sin(Φ) dρdΦdθ. Remember that (ρ^2) sin(Φ) is always included when calculating spherical triple integral. When determining the limits for the integral begin from the outside in, thus beginning with θ. θ represents the angle of counterclockwise rotation around the z-axis. Since it is a sphere, 0<θ<2π for the limits. For the Φ limits, it is angle from the +z-axis to the -z-axis. In this case you begin at the cone, π/4, and go until where the sphere ends, the xy-plane which is π/2, thus π/4<Φ<π/2. Finally, ρ is the radius of the object. Since this object is centered at the origin, its lower limit is 0. The upper limit is given by ρ= 4 cos(Φ), thus 0<ρ<4 cos(Φ). Once setup, evaluate as usual by treating different variables as constants when needed. Let me know if you need help evaluating!

Chemical Engineering
TutorMe
Question:

Could you please explain to me the concept of entropy and how it applies to chemical processes as well as how this might help some a homework or exam problem?

Jeremiah B.
Answer:

Sure! Simply put, entropy is a measurement of disorder. Entropy can increase (ex. temperature increasing). Now in regards to chemical engineering, processes are sometimes assumed to be reversible which means there is no change in entropy of the process or device. This however is an ideal, unattainable situation because it assumes unrealistic properties of the system like a frictionless environment. This hypothetical can be applied to the real-world by allowing us to set maximum to devices, like maximum amount of work that can be produced by a turbine. Typically for problems, no change of entropy is assumed which allows for a connection between the beginning and end of the process, meaning the same entropy value can be used when looking for data in tables. Hope this helps!

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