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Tutor profile: Jenna P.

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Jenna P.
Second year medical student. Tutor for 5 years, math and science
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Subject: Organic Chemistry

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Question:

Predict the position of equilibrium for the following reaction: NH2 + O <------> -NH + O | || | || / \ / \ / \ / \ O- OH

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Jenna P.
Answer:

. NH2 + O <------> -NH + O | || | || / \ / \ / \ / \ O- OH Recall: equilibrium basically means the two sides are in a bit of a competition for H+ because you have a weak base on each side. The side of the equilibrium that has the base which allows the negative charge to be MORE STABLE will be the side of the reaction that is favoured. So, we have a comparison on our hands....which negative charge is more stable? O -NH || | / \ / \ O- We have 4 factors that we must compare in order to determine stability, listed here in order of importance: 1) ATOM: -Electronegativity: which atom is more electronegative i.e. which atom LIKES having negative charge and thus is more comfortable having negative charge around it? -Size: which atom is larger i.e. has more space to distribute that charge around? 2) RESONANCE: -Is there a resonance structure that spreads the charge over more than one atom? 3) INDUCTION: -Are there electronegative atoms around that draw negative charge away from the atom holding negative charge i.e. making it LESS negative and more stable? -Are there electron donating atoms around that actually contribute more negative charge to the atom holding negative charge i.e. making it MORE negative and less stable? 4) ORBITAL: -Remember in terms of stability, sp > sp2 > sp3 Why? It helps to first Google the comparative sizes of the orbitals for a visual, but essentially the sizes are sp = smallest, sp2 = medium, sp3 = largest. You might think that the larger orbital space means that the negative charge is able to be distributed over a larger area and thus is more stable. BUT this is incorrect. Actually, the smaller orbital spaces are MORE stable than the larger ones because they hold the electrons closer to the nucleus which has + charge in it, and thus MORE STABLE by having the negative charges held closer to positive charge. So....back to comparing our two bases. O -NH || | / \ / \ O- 1) ATOM: -The first compound has the negative charge on an O which is more electronegative than N, where the negative charge is placed in the second compound RESULT? First compound is more stable in terms of atom. 2) RESONANCE: -The first compound can be stabilized by resonance. (It's a bit difficult to show with a keyboard here but you can push the double bond up onto the oxygen on top and then bring the negative charge from the bottom oxygen into the double bond position) -The second compound cannot be stabilized by resonance (There's nowhere to push the negative charge...) RESULT? First compound is more stable in terms of resonance. At this point, we could already say that the first compound is favoured over the second in terms of stabilizing negative charge. In fact, we probably could have said this after the first step of looking at which atom the negative charge was on, but checking the other steps to be certain isn't a bad idea. For completeness sake, let's do step 3 and 4 anyway. 3) INDUCTION: -There really are no significant electron donating or electronegative groups playing a role in either molecule 4) ORBITALS: -Both are on sp3 orbitals, so there is also no difference between either molecule Thus, the first molecule is able to stabilize negative charge more than the second. As this molecule is on the left side of the equilibrium, it is the left side of the equilibrium that is favoured!

Subject: Basic Chemistry

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Question:

Find the theoretical yield of the reaction below, given 24.0g of each reactant: 3Mg + 1N2 --> 1Mg3N2

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Jenna P.
Answer:

A "theoretical yield" is essentially the amount of product that we SHOULD get based on calculations from the original starting amounts of reactant given. We need to use stoichiometry to find how much product we should end up with. 3Mg + 1N2 --> 1Mg3N2 24.0g 24.0g ? However, we have 2 reactants that we are dealing with. How do we know which one to use for our calculation? We must figure out which reactant is our LIMITING REAGENT. Essentially, this means finding which reactant is going to RUN OUT FIRST and thus LIMIT the reaction from continuing to take place. We know we are starting with the same amount of grams of each reactant (24.0g). But remember, each chemical species has a different molar mass i.e. a different number of g/mol. Recall that moles gives us a measure of the number of ATOMS that we have (1 mole = 6.02214179 × 10^23 atoms). So, if we have molar masses of 2g/mol for one substance, and 3g/mol for another substance, even if we have the same amount of grams for each substance, we're going to have a different number of moles i.e. a different number of atoms of each substance. We compare things in moles so that we can compare the number of ATOMS of each species involved because it's the atoms reacting with each other and causing the reaction and thus when we run out of atoms of one substance, the reaction will be limited. SO.........our next step is to convert our reactants from grams to moles. 3Mg + 1N2 --> 1Mg3N2 24.0g 24.0g ? /24.3g/mol /14.0g/mol =0.9877mol =1.7142mol So looking at our moles, it may be easy to jump to the conclusion that since Mg has the lesser number of moles, it will be the limiting reagent. It might be, but we can't just look at number of moles to see what limits the reaction because the reaction proceeds depending on STOICHIOMETRIC RATIOS. So here, for every 3 atoms of Mg used, 1 atom of N2 is used.... We have 0.9877mol of Mg...so when multiplied by the stoichiometric ratio we just stated, we multiply the 0.9877mol of Mg by 1/3 to find the number of moles of N2 that would be needed if all of the moles of Mg were to be used up: 0.9877mol x 1/3 = 0.3292mol So, if we were to use up all of the 0.9877mol of Mg, we would need to have 0.329mol of N2 for the reaction to proceed. Let's look back up at how many moles of N2 we actually have: 1.7142mol....so we have plenty of moles of N2 for all of the Mg to be used up. Thus, N2 is NOT our limiting reagent! For completeness sake, you can do the calculation the other way to see why Mg IS our limiting reagent: 1.7142mol N2 x 3/1 = 5.142mol of Mg needed But we only have 0.9877 moles of Mg SO, if we were to use all of the N2, we wouldn't have enough Mg, and thus Mg is the limiting reagent! Now that we have our limiting reagent, we know which reactant to use in our calculation to calculate our yield of Mg3N2....which is Mg! 3Mg + 1N2 --> 1Mg3N2 24.0g 24.0g ? /24.3g/mol /14.0g/mol =0.9877mol =1.7142mol -----------------------------------------------x1/3--------> 0.3292mol We have our number of moles of Mg3N2 produced which is 0.3293mol! But, this isn't very useful for lab purposes. Scales measure grams, so we need to convert back into grams. Remember, we need to now MULTIPLY by molar mass to get our grams from moles -if you forget whether you need to divide or multiply to get grams or moles, just look at the UNITS use whichever one gets the units that you don't need to cancel out... ex) 2g / 3g/mol ----> the grams cancel out because you are dividing the grams. This leaves you with moles ex) 5mol x 3g/mol -----> the moles cancel out because you are dividing the moles. This leaves you with grams So.... 0.3292mol x _____g/mol of Mg3N2 Figure out Mg3N2 molar mass: (3 x 24.305g/mol) + (2 x 14.001g/mol) = 100.917g/mol 0.3293mol x 100.917g/mol = 32.232g We have found our theoretical yield: 32.232g of Mg3N2

Subject: Algebra

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Question:

Factor the algebraic expression 16y^2 − 25x^2 + 10x − 1

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Jenna P.
Answer:

The first thing we want to notice here is that the latter 3 terms of the equation make up what we call a "quadratic" expression. So let's first review how we factor quadratic expressions: If we have x^2 + 3x - 4, we know that the format of our factored expression has to be: (x + __ ) (x + __ ) ----- with FOIL, this gives us x^2 when we multiply both x's together So, how do we figure out what to fill in the blanks with? Well, let's write it another way (x + a) + (x + b) When we FOIL that out, we get x^2 + ax + bx + ab When we combine like terms, we get x^2 + (a + b)x + ab In other words, we need the product of a and b to give us -4 and the sum of a and b to give us 3: a x b = -4 a + b = 3 If we just look at the a x b for now, what are possible numbers here to give us a product of 4? -2 x 2 = 4 -1 x 4 = 4 Now, we look at the sum of the number combos we just wrote down to find a sum of 3: 2 + 2 = 0 -1 + 4 = 3 --- this is the combination of numbers that we want then So, we fill in the equation: (x + 4) (x - 1) {To double check your work, FOIL the expression back out} NOW....let's apply that to our piece of the original expression that is a quadratic: 16y^2 - (25x^2 - 10x + 1) ---- we're going to separate the quadratic piece from the rest of the expression and take the negative sign outside of our quadratic piece to make it easier to deal with. Thus, we are left with: 25x^2 - 10x + 1 Again, to factor a quadratic, we know we need to separate into two binomials in the following form: (x + a) (x + b) Since there is a 25 in front of the x^2 and 25 is a perfect square, we put it's square root in front of our x's as well: (5x + a) (5x + b) Then we need to find our a and b that will give us a product of 1 and a sum of -10. But, we can't forget about the coefficient of 5 in front of the x's --- those are going to be multiplied with the a and b term as well to be the final coefficients in front of x when we do the "O" and "I" part of FOIL. a x b = 1 5a + 5b = -10 What are possible number combos that give us a product of 1? -1 x -1 = 1 1 x 1 = 1 When we substitute these into 5a + 5b, it is -1 and -1 which give us a sum of -10. So, the combo of numbers we want here is -1 and -1. Let's substitute those into our binomials: (5x - 1) (5x - 1) which can also be written as (5x - 1)^2 We are done with that first quadratic part of the original expression! We are then left with: 16y^2 − (5x - 1)^2 Now let's look at the 16y^2. This is simply (4y) (4y), which is also written as (4y)^2. We are then left with: =(4y)^2 - (5x - 1)^2 This is actually the same format as a^2 - b^2 {where a = 4y and b = (5x - 1)}. We know that factoring out an expression of a^2 - b^2 into two binomials must take the form of (a + b) (a - b) Therefore: a^2 - b^2 = (a + b) (a - b) (4y)^2 - (5x - 1)^2 = {4y + (5x - 1)} - {4y - (5x - 1)} We can then distribute the + and - to get: {4y + 5x + 1} - {4y - 5x + 1} And that's our final answer!

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