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Tutor profile: Khowa A.

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Khowa A.
Chemical Engineering Graduate; Biotechnology and Pharmaceutical Engineering Researcher
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Questions

Subject: Chemistry

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Question:

The incomplete combustion of ethane (C2H6) yields carbon dioxide, carbon monoxide, and water. Write out this combustion equation and balance it. If there are 200g of oxygen and 40g of ethane, how many grams of carbon monoxide are produced if the reaction goes to completion?

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Khowa A.
Answer:

Step 1: Write out the equation without coefficients C2H6 + O2 --> CO2 + CO + H2O Step 2: Balance in the following order CHO C2H6 + O2 --> CO2 + CO + H2O (C's are balanced) C2H6 + O2 --> CO2 + CO + 3H2O (C's & H's are balanced) C2H6 + 3O2 --> CO2 + CO + 3H2O (C's, H's & O's are balanced) Step 3: Determine the limiting reagent. (You will need a periodic table for this) -To determine the limiting reagent, you need to determine how many reactions can be run with the amount of each chemical specified (calculations should be done in moles). -Molar mass of O2 = 32 g/mol -Molar mass of C2H6 = 30.08 g/mol -Molar mass of CO = 28.01 g/mol -Next, we do some unit analysis using the molar ratios in the balanced equation -200g O2 * (1 mol O2/32g O2) * (rxn/3 mol O2) = 2.0883 rxn -40g C2H6 * (1 mol C2H6/30.08g C2H6) * (rxn/1 mol C2H6) = 1.3298 rxn -This means that we have enough oxygen for over 2 reactions but only enough ethane for a little more than one reaction. Therefore ethane is our limiting reagent. All stoichiometric calculations must be done using the limiting reagent Step 4: Use stoichiometry to determine the mass of CO produced 40g C2H6 * (1 mol C2H6/30.08g C2H6) * (1 mol CO/1 mol C2H6) * (28.01g CO/1 mol CO) = (40*28.01)/30.08 = 37.25g CO

Subject: Calculus

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Question:

Sketch a graph of the curve denoted by the equation: x(2x^2-3x)-3

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Khowa A.
Answer:

To sketch a curve, you must first expand the equation: x(2x^2-3x)-3 = 2x^3 - 3x^2 - 3 Looking at this equation, you can see that it is a trinomial (polynomial of the third degree) with a positive cubic coefficient (2). Therefore, you know that the graph will have two minima/maxima and will ascend in the positive x and y directions. Next, find the first derivative of the function: f' = 6x^2 - 6x Setting this first derivative to zero and solving the equation will show the location of the minimum and maximum points of the graph. 0 = 6x^2 - 6x; 6x(x-1) = 0; x=0 & x=1 Now that we know that 0 and one are the points of the maximum and minimum, you can plug in those x values into the original function to find the corresponding y values: for x = 0, y = -3 & for x = 1, y = -4 Now you have two points on the graph. Next, find the second derivative of the function. That is, the derivative of the first derivative: f'' = (f')' = 12x - 6 We will use the second derivative to find the inflection points on the curve. An inflection point is where the concavity (the curvature) of a graph changes. Set the second derivative to zero and solve for x: 0 = 12x - 6; x = 1/2 Now we know that there is a change in curvature at x = 1/2, y = -3.5 To find how this curvature changes, we will test x values below and above the inflection point. Let's take -1 and 1 as our test points: -for x = -1, the second derivative of the function is negative. This means that the curve is opened downwards. -for x = 1, the second derivative of the function is positive. This means that the curve is opened upwards. Combining the information from the first and second derivatives of the function and from the general knowledge about third degree polynomials, we can start our curve drawing: The curve starts at negative infinity on the y axis, then comes up to a maximum at (0,-3). The curve then drops and changes inflection at (1/2.-3.5) and continues to a minimum at (1,-4). Finally, the curve continues up to positive infinity on the y axis. Apart from the specifically known x-axis coordinates, it is not important to get the x scale of the graph super accurate.

Subject: Algebra

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Question:

Tommy and Bridgette are conducting a science experiment. They are given a 8L jug which contains 5L of 25% acetic acid. If they are to add 10% acetic acid to the jug (while using all of the 25% solution), will the 8L jug be large enough to make a new 15% solution?

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Khowa A.
Answer:

No. They will need 10L of 10% acetic acid to be added to the 5L of 25% acid to make a 15L of a 15% solution. They will need a 15L jug to make the solution. Work: 5L of 25% solution = (0.25*5)L of acetic acid xL of 10% solution = (0.1*x)L of acetic acid (5+x)L of a 15% solution = (0.15*(5+x))L of acetic acid The first two amounts of acetic acid have to add up to the last amount because those are the only sources of the acid. We can assume that the acid is being diluted with water, but that is irrelevant for this solution. Thus, (0.25*5) + (0.1*x) = (0.15*(5+x)) Then, isolating for x gives: 1.25 + 0.1x = 0.75 + 0.15x; x = 10L

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