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# Tutor profile: Trent A.

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Trent A.
Almost completed PhD in Chemistry with B.A.s in Chemistry, Physics, and Math
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## Questions

### Subject:Chemistry

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Question:

The compound 2,3-diaminothiophene (molecular formula SC4H2(NH2)2) reacts at a 1:1 ratio with 2,3-butadione (molecular formula C4H6O2) to yield one equivalence of 2,3-dimethylthieno[2,3-b]pyrazine (molecular formula SC6N2H2(CH3)2) and two equivalences of water. If 1.00 grams of 2,3-diaminothiophene reacted with 1.00 grams of 2,3-butadione, what is the theoretical yield in grams? If 0.90 grams of 2,3-dimethylthieno[2,3-b]pyrazine was isolated, what was the percent yield? Using LeChatlier's principle, explain if you would expect the yield to increase or decrease if water was added to the reaction once it reached equilibrium.

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Trent A.

There are a lot of words in this problem that set the stage for it, but do not help with the solving of it. This is a situation where we have two reagents reacting and giving a product with a lot of words around the essential information. The first thing that needs to be done is the limiting agent needs to be determined. To do that, we need to determine the molecular weights of the two reagents: 2,3-diaminothiophene and 2,3-butadione. To determine the molecular weight of each, we need to know its composition. The reagent 2,3-diaminothiophene has one sulfur, four carbons,six hydrogens (two from each of the amines) and two nitrogens. This gives us a molecular weight of (1*32.06 + 4 * 12.011 + 6 * 1.01 + 2*14.01)=114.18 grams/mol. The same can be done for 2,3-butadione, which has four carbons, six hydrogens, and two oxygens. This gives us a molecular weight of (4*12.011 + 6 * 1.01 + 2 * 16.00) = 86.10 g/mol. We can now determine how many mols of each reagent we have. To do this, we take the mass that we have and divide it by the molecular weight. For 2,3-diaminothiophene that means we have (1.00 g / 114.18 g/mol) 8.759 mmol (0.008759 mols). For 2,3-butadione we have (1.00 g/86.10 g/mol) 11.61 mmol (0.01161 mols). The two react at a 1:1 ratio, so whichever one there is less of will be the limiting reagent. That means that 2,3-diaminothiophene is the limiting reagent. The ratio between 2,3-diaminothiophene and 2,3-dimethylthieno[3,4-b]pyrazine are at a 1:1 ratio, so the number of mols of the limiting reagent will be the number of mols for the theoretical yield (8.759 mmol). The question asks for the theoretical yield in grams, so there is still the need to convert the mols to grams. To do this we need to determine the molecular weight of the product. It contains one sulfur, eight carbons, two nitrogens, and eight hydrogens, giving it a molecular weight of (1*32.06 + 8*12.011 + 2*14.01 + 8 * 1.01) 164.25 g/mol. To get the mass we now multiply the mols by the molecular weight to get (0.008759 mols * 164.25 g/mol) 1.439 grams. However, we are limited to three significant figures, giving us a theoretical yield of 1.44 grams. The second part of the question was to determine the percent yield if only 0.90 grams was isolated. To calculate this we divide the actual yield by the theoretical yield (0.90 g / 1.44 g * 100%) to get 62.5%. However, because the isolated mass only has two significant figures, so out final answer is actually 63%. The discussion part of the question asks about water and LeChatlier's principle. Water is one of the products in the reaction, so to add more water to it acts as a stressor that must be brought into balance. The increase in a product will result in a shift to the left (or to the reactants). This means there will be a decrease in the yield of the reaction.

### Subject:Calculus

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Question:

What is the value of "j" such that the two integrals: integral(1/2*x-1/2 dx, 0, j) and integral(3 dx, j, 2) are equal to each other and j is between 0 and 2?

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Trent A.

In order to determine what value of "j" gives equal values for these integrals, we must first evaluate the integrals. The integral of the first gives us x^2 - 1/2*x, evaluated from 0 to j. Evaluating this gives us a final value of j^2-1/2*j. Remember that we do not need to include the +c for a definite integral. The second integral must also be solve. The second integral gives us 3x, evaluated from j to 2. Evaluating this gives us 6-3j. Setting these two values equal to each other gives us the following: j^2-1/2*j = 6 - 3j [This can be rearranged to a quadratic equation] j^2 + 2.5*j - 6=0 [This can be factored to (j+4)(j-1.5)] (j+4)(j-1.5)=0 [If you are not able to identify the factors of this equation, the quadratic formula would also work for finding the solutions] This final equation gives us j as a value of -4 and 1.5. However, the question has the restriction that j must be between 0 and 2, so our final answer is j=1.5.

### Subject:Physics

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Question:

Most baseball pitchers can throw a fastball 90. miles per hour. If home plate is 60. feet away from the pitcher's mound, how long will it take for the fastball to travel from the pitcher to homeplate? If a batter can swing a bat in 0.15 seconds, how far will the ball have traveled when the batter starts their swing in order to hit a fastball?

Inactive
Trent A.

The first step is to identify the best equation to determine the time of the ball traveling. This starts with distance = velocity * time. We have a distance of 60. feet and a velocity of 90. miles per hour. Unfortunately we have a units issue so we need to convert miles per hour to feed per second. 90 miles/hour *(5280 ft/1 mi)*(1 hour/3600 seconds)=132 ft/sec. Substituting this into the earlier equation gives us 60 ft = 132 ft/sec*time. All of the units are the same and we can do some algebra to get our time: time=60 ft / (132 ft/sec) = 0.45 seconds. The second half of the problem uses the results from the first half. If the ball travels a for a total of 0.45 seconds and the batter must swing 0.15 seconds before the ball gets to the plate, the ball is only going to travel 0.45 sec - 0.15 sec = 0.30 seconds before the batter must swing. The distance it travels in 0.30 seconds (using distance=velocity*time) is: distance=132 ft/sec * 0.30 sec = 39.6 feet, which rounds to 40 feet with appropriate significant figures.

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