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# Tutor profile: Nick G.

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Nick G.
Engineering Grad with Physics and Chemistry Background
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## Questions

### Subject:Organic Chemistry

TutorMe
Question:

Why can Chloro-trimethyl Methane not engage in an Sn2 Reaction under basic condition to make tert-butanol?

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Nick G.

Draw the 3-D structure. Build it as well if you have a modeling kit. Now think of the electronics here. The chlorine makes the C electron poor, a prime candidate for Sn2 reactions. So why no reaction? From the structure, we see a lot of sterics happening here. By sterics, I mean that the C center has too much traffic from the other atoms to allow another atom to attack. So the sterics are the reason here.

### Subject:Physics

TutorMe
Question:

Electrode is a lesser known electric pokemon in the shape of a perfect sphere. From bulbapedia, an Electrode has a body height of 1.2 m and weight of 66.6 kg. Using this, how much weight could an electrode theoretically hold in water before sinking?

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Nick G.

Electrode is a sphere, so its height is equivalent to the diameter. This lets us find the volume as: V = 4/3*PI*(D/2)^3 = 4/3*(3.14159)*(1.2 m /2)^3 = 0.91 m^3 The equivalent volume of water weighs: mass water = 0.91 m^3 * (1000 kg / m^3) = 910 kg Based solely on buoyancy, Electrode could hold : (mass water displaced - mass electrode) = 910 kg - 66.6 kg = 843 kg The weight would be: 843kg*9.81m/s^2 = 8273 N (0.22 lbs/N) = 1820 lbs And yet it can't use surf...

### Subject:Basic Chemistry

TutorMe
Question:

Oxygen and Hydrogen are used in a fuel cell to produce electricity and water. If one feeds 0.5 L of hydrogen to the fuel cell, how much water could come out max? How much oxygen needs to be fed, minimum, for this amount of water? Assume hydrogen is fed at STP.

Inactive
Nick G.

Start by writing the basic reaction: O2 + 2H2 -> 2H2O Next, use the ideal gas law to find hydrogen's molar density. R is ~8.3 J/mol K ~ 0.083 atm L / mol K. PV=nRT => n/V = P/RT => n/V = (1atm)(0.083 atm L/mol K)^-1 (273K)^-1 n/V = 0.0441 mol/L Now we can find the moles of hydrogen, n: n= (n/V)(V) = (0.0441 mol/L)(0.5L) n = 0.0221 mol Finally, we can use the reaction stoichiometry to find the mols of water. The ratio of hydrogen to water here is 2 to 2, or 1 to 1: n = 0.0221 mol H2 (2 mol H2O/2 mol H2) = 0.0221 mol water To find the oxygen needed, again we use the ratio from the reaction. This reaction requires 1 mol of O2 per 2 mols of H2O. Therefore: nO2 = 0.0221 mol water (1 mol O2/2mol H2O) = 0.0111 mol O2 Note: I round my numbers to the 2 sig figs. In some science classes, you will be asked to round to more sig figs. Always follow the directions for the given problem.

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