Tutor profile: Jenna L.
300.0 grams of a compound composed of carbon, hydrogen and oxygen atoms is burned in an excess of O2 to form 349.5 grams of CO2 and 125.2 grams of H2O. Find the empirical formula of the compound.
In order to find the empirical formula, the ratio of moles of carbon, hydrogen, and oxygen must be determined. Start with carbon and hydrogen, as using the grams of CO2 and H2O given, the number of moles of carbon and hydrogen in the original compound can be determined. For carbon, convert grams CO2 to moles CO2. The molecular weight of CO2 is 44.01 g/mol, so the moles of CO2 in 349.5 grams is 7.941 moles. In each mole of CO2, there is one mole of carbon, so the moles of carbon in the original compound is also 7.941 moles. For hydrogen, follow a similar process. Convert grams H2O to moles H2O. The molecular weight of H2O is 18.02 g/mol, so the moles of H2O in 125.2 grams is 6.948 moles. In each mole of H2O, there are 2 moles of hydrogen. Therefore the number of moles of hydrogen in the original compound is 13.90 moles. To find the empirical formula, it is also necessary to find the number of moles of oxygen. However, a different method must be used because the compound is burned in an excess of O2. Therefore, the oxygen in the CO2 and H2O is from both the compound and the O2. Instead, convert the moles of carbon and hydrogen in the compound to grams. By subtracting the grams of carbon and hydrogen from the total mass of the compound, the mass of oxygen will be obtained. The molecular weight of carbon is 12.01g/mol and the molecular weight of hydrogen is 1.01 g/mol. Therefore 7.941 moles of carbon is 95.38 g, and 13.90 moles of hydrogen is 14.03 g. The mass of oxygen in the compound is thus 300.0 g - 95.38 g - 14.03 g = 190.6 g. The molecular weight of oxygen is 16.00 g/mol, so the moles of oxygen from 190.6 g is 11.91 moles. Now that the values of moles for each of the elements is found (C = 7.941 mol, H = 13.90 mol, O = 11.91) the ratios between the moles can be determined. Divide each of the numbers of moles by the element with the smallest number of moles, in this case carbon with 7.941. The resulting values will be 1 C/C, 1.750 H/C, and 1.500 O/C. Multiply these three numbers by an integer so that all three numbers are integers. In this case, multiply each by 4 to get 4 C, 7 H, and 6 O. These numbers are the numbers of atoms in the empirical formula, so the result is C4H7O6.
Find the area of the region where r = 3 and r = 3cos(θ) + 3 overlap.
The first step in this problem is to determine where the two curves intercept. Set r = 3 equal to r = 3cos(θ) + 3 to solve for θ. To solve 3 = 3cos(θ) + 3, divide both sides by 3. Next, 1 = cos(θ) + 1, subtract 1 from both sides. This will yield 0 = cos(θ). cos(θ) can equal 0 when θ = π/2, - π/2, or 3π/2. In order to calculate the area of the overlapping region, it must be determined which curve is inside the other. Consider the graph of the two functions. The curves intersect twice. On the interval (-π/2, π/2), r = 3 is less than r = 3cos(θ) + 3. On the interval (π/2,3π/2), r = 3cos(θ) + 3 is greater than r = 3. Remember that no further intervals need to be considered, as for the purpose of area, the length of the intervals should not be longer than 2π. This problem will need two integrals, as on the first interval r = 3 is the inner curve and on the second r = 3cos(θ) + 3 is the inner curve. The first interval should be 1/2 times the integral of 3^2 (dθ) with bounds from -π/2 to π/2, and the second should be 1/2 times the integral of (3cos(θ) + 3)^2 (dθ) with bounds from π/2 to 3π/2. In order to make this explanation clearer, one interval will be evaluated at a time. Simplify the first integral to 1/2 * int(9 dθ) with bounds -π/2 to π/2 (where "int" stands for "integral"). Integrate to get 1/2 * (9θ) from -π/2 to π/2. Plug in to get 1/2 * (9π/2 - -9π/2) and simplify to get 9π/2. For the second integral, simplify to get 1/2 * int(9cos(θ)^2 + 18cos(θ) + 9)(dθ) with bounds π/2 to 3π/2. In order to integrate 9cos(θ)^2, use the trigonometric property cos(θ)^2 = 1/2(cos(2θ) + 1). Rewrite the integral as 1/2 * int(9/2(cos(2θ) + 1) + 18cos(θ) +9)(dθ) with bounds π/2 to 3π/2. Integrate to get 1/2 * (9/2(1/2sin(2θ) + θ) + 18sin(θ) + 9θ) from π/2 to 3π/2. Plug in to get 1/2 * (9/2(1/2sin(3π) + 3π/2) + 18sin(3π/2) + 27π/2 - 9/2(1/2sin(π) + π/2) - 18sin(π/2) - 9π/2). Simplify to get 1/2 * (9/2 * 3π/2 - 18 + 27π/2 - 9/2 * π/2 - 18 -9π/2) and simplify again to get 27π/2 - 36. To get the final area, add the two integrals together. 9π/2 + 27π/2 - 36 equals 18π - 36, which is the area of the overlapping region.
Find the volume of a solid by revolving the region bounded by the graphs of f(x) = x^2 and g(x) = sqrt(x) about the y-axis. (note that sqrt stands for "square root")
The first step to solve this problem is to get a good idea of what these functions look like graphically. In order to set up the integral needed to add up the volume, it is necessary to have a clear idea of the functions. f(x) and g(x) intersect only twice, at (0,0) and (1,1). Consider the axis of rotation, which in this problem is the y axis. Since the area will be rotated about the y-axis, it is best to work with the functions in terms of y. This will give us f(y) = sqrt(y) and g(y) = y^2. There is a gap between the area enclosed by f(y) and g(y), and the axis of rotation. Because of this gap, the final volume will be the volume of the outer function rotated about the y-axis minus the volume of the inner function rotated about the y-axis. Thinking graphically, decide which function is closer to the y-axis. In this case, g(y) = y^2 is closer. To find the volume of the solid, multiply 𝜋 by the integral of the outer function f(y) squared subtracted by the inner function g(y) squared. The bounds of this integral are the intersections of f(y) and g(y), which are y = 0 and y = 1. The inside of the integral will be (sqrt(y))^2-(y^2)^2(dy). Simplify this to get y-y^4(dy). Integrate to get (y^2)/2-(y^5)/5 from 0 to 1. Plug in the bounds to get ½-⅕, which simplifies to 3/10. Finally, multiply by 𝜋 to get the volume of 3𝜋/10.
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