# Tutor profile: Eddie W.

## Questions

### Subject: Chemistry

When I worked as a peer learning facilitator for general chemistry, I came up with some fun, in-depth questions drawing on multiple concepts. Here's one of them: Liquid formic acid, CH2O2 (l), is found naturally in some species of ants and other insects, and it has practical use for humans as a preservative and antibacterial agent. The carabid beetle produces formic acid at a rate of 126 µg/day [1]. The combustion reaction for formic acid is shown below: $$ CH_2 O_2 (l)+\frac{1}{2} O_2 (g) → CO_2 (g)+H_2 O(g),∆H°combustion= -255 kJ/mol$$ (1) You have a cup containing 6 fluid ounces of coffee, but it is at room temperature (77 °F)! You want to heat it to 130 °F, your optimal coffee-drinking temperature, by burning formic acid. How many years must one carabid beetle work to heat your beverage? Answer to the nearest tenth of a year. Assume constant pressure, and use the heat capacity ($$c = 4.18 \frac{J}{g*K}$$) and density ($$1g/mL$$) of water as an estimate for the coffee, assuming 100% of the combustion energy goes toward heating the beverage. Useful: $$ T(°C) = (T(°F) – 32)*(5/9), ∆H= q_p, q=mc∆T, 1 fluid ounce = 29.5735 mL, 1 year = 365.2564 days, MW formic_.acid = 46.025 g/mol $$. Reference: [1] Rossini, Carmen & Attygalle, Athula & González Ritzel, Andrés & R. Smedley, Scott & Eisner, Maria & Meinwald, Jerrold & Eisner, Thomas. (1997). Defensive production of formic acid (80%) by a carabid beetle (Galerita lecontei). Proceedings of the National Academy of Sciences of the United States of America. 94. 6792-7. 10.1073/pnas.94.13.6792.

Solution (1) Convert the initial and final temperatures of the coffee to Celsius: $$ T(°C) = (T(°F) – 32)*(5/9)$$ (2) $$ T_i (°C) = ((77) – 32)*(5/9) =25 °C $$ (3) $$ T_f (°C) = ((130)– 32)*(5/9)=54.4°C $$ (4) (2) Find ∆T: $$ ∆T= T_f-T_i=(54.4 -25)°C=29.4°C$$ (5) (3) Find the estimated mass of the coffee using the given volume and density: $$6 oz*\frac{29.5735 mL}{oz}*\frac{1g}{mL}=177.441 g $$ (6) (4) Use q=mc∆T to find the required energy to heat the beverage, with the mass found in equation (6), the given heat capacity, c, and the ∆T found in equation (5), realizing that 1 degree Kelvin is the same interval as 1 degree Celsius: $$q=mc∆T=(177.441g)(4.18 J/(g*K))(29.4 K)=21839 J $$ (7) (5) Realizing that $$∆H= q_p$$, and assuming 100% efficiency, it is possible to use $$∆H°combustion$$ as a conversion factor to get to moles of formic acid: $$ 21839 J*\frac{1 kJ}{(1000 J}*\frac{1mol}{255 kJ}=8.5643*10^{-2} mol CH_2 O_2 (l) $$ (8) (6) Using the moles obtained in equation (8), perform dimensional analysis, converting from moles to grams using the molecular weight (MW formic acid = $$46.025 g/mol$$), then from grams to micrograms, then from micrograms to days, given the conversion factor of 126 µg/day, and finally from days to years (1 year = 365.2564 days): $$8.5643*10^{-2} mol*\frac{46.025 g}{mol}*\frac{10^6 μg}{g}*\frac{1 day}{126 μg}*\frac{1 year}{365.2564 days} $$ (9) (7) Solve the above, and find that it would take one carabid beetle about 85.6 years to heat your coffee by the indicated amount!

### Subject: Statistics

Here's an example statistics question I answered on Quora: How many ten-digit telephone numbers are possible if the first three digits must be different?

Note: this answer assumes all digits 0–9 are allowed, i.e. that the only stipulation is difference among the first three numbers. For this problem, the first three digits can be treated as distinguishable objects; imagine that you have ten different colored balls to place into three slots, and you want to know all the permutations for this. The formula for permuting $$n$$ distinguishable objects in $$r$$ slots is given by: $$_nP_r = \frac{n!}{(n-r)!}$$ In this case, for the first three digits, $$n = 10$$, $$r = 3$$: $$_{10}P_3 = \frac{10!}{(10-3)!} = 10*9*8 = 720$$ For the remaining seven digits, there are $10^7$ possibilities. Each allowed permutation of the first three digits contains all the possibilities for the remaining seven digits, so the final answer is: $$_{10}P_3 * 10^7 = 720 * 10^7 = {7.2 * 10^9}$$ $$possible$$ $$numbers$$

### Subject: Algebra

A great "final exam" for mastery of algebra I is proving the quadratic formula without help. Can you prove it?

Standard form: $$ ax^2 + bx + c = 0 $$ Divide by a: $$ x^2 + \frac{bx}{a}+\frac{c}{a} = 0 $$ Subtract c/a: $$ x^2 + \frac{bx}{a} = -\frac{c}{a} $$ Complete the square: $$ x^2 + \frac{bx}{a} + ( \frac{b}{2a})^2 = -\frac{c}{a} + ( \frac{b}{2a})^2 $$ Factor: $$ (x+\frac{b}{2a})^2=-\frac{c}{a} + ( \frac{b}{2a})^2 $$ Simplify right side: $$ (x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2} $$ Take square root: $$ x+\frac{b}{2a} = \pm \sqrt{\frac{b^2-4ac}{4a^2}} $$ Subtract b/(2a): $$ x = -\frac{b}{2a}\pm \sqrt{\frac{b^2-4ac}{4a^2}} $$ Simplify: $$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$

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