QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.2 : (a) Let | - > = 1 | x > + 0 | y >, | | > = 0 | x > + 1 | y >. Find the value of 2 | x > + 3 | y > in term of | - > and | | >. (b) Let m to be the reduced mass. Find the value of m in term of Ma and Mb where 1 / m = 1 / Ma + 1 / Mb.
QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - ANSWER 31.2 : (a) For | - > = 1 | x > + 0 | y >, multiply it with 2 to produce 2 | - > = 2 | x > + 0 | y >, then 2 | x > = 2 | - > - 0 | y > as first equation. For | | > = 0 | x > + 1 | y >, multiply it with 3 to produce 3 | | > = 0 | x > + 3 | y >, then 3 | y > = 3 | | > - 0 | x > as second equation. Finally first equation plus second equation to produce 2 | x > + 3 | y > = 2 | - > + 3 | | >. (b) Let 1 / m = 1 / Ma + 1 / Mb = (Ma + Mb) / (Ma x Mb). Then m = (Ma x Mb) / (Ma + Mb). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; email@example.com; http://kangchuentat.wordpress.com.
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Question 60 – During the landing process of an airplane, the velocity is constant at v. (a) If the displacement of the plane is x at time t, find the differential equation that relates t, x and v. (b) The plane has 2 parts of wheels – the front and the back, separated by a distance L. The front part of the wheel touches the land first, that allows the straight body of the plane to form an angle T with the horizontal land. If the vertical distance between the back part of the wheel and the horizontal land is y, find the equation of y as a function of L and T. (c) Find the differential equation that relates dy as a function of dt, v and sin T. (d) Find the differential equation that consist of dy as a function of y, L, v and dt. (e) Find the equation of y as a function of v, L, t and C where C is a constant. (f) When t = 0, prove that y = exp C as the initial value of y.
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CHEMICAL FLUID MECHANIC - EXAMPLE 3.2 : The terminal velocity of a falling object, v is given by v = sqrt [ 4g (R - r) D / (3Cr) ] where sqrt is the square root of, g = 9.81, D = 0.000208, R = 1800, r = 994.6, m = 0.000893. The Reynold number, L is given by L = rD (v) / m. The C for various conditions are : C = 24 / L for L < 0.1; C = 24 (1 + 0.14 L^0.7) / L for 0.1 <= L <= 1000; C = 0.44 for 1000 < L <= 350000; C = 0.19 - 80000 / L for 350000 < L. Find the value of v for the situation above by trial and error, ^ is power, <= is less than or equal to.
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Question 75 - The molecular weights M in kg / mol of 3 different monomers a, b and c in a polymer are Ma = 14, Mb = 16 and Mc = 18. The fraction of polymer chain X of 3 different monomers a, b and c in a polymer are Xa = 0.5, Xb = 0.3 and Xc = 0.2. (i) Calculate number average molecular weight by using the formula Ma Xa + Mb Xb + Mc Xc. (ii) Calculate weight average molecular weight by using the formula (Ma Xa Ma + Mb Xb Mb + Mc Xc Mc) / (Ma Xa + Mb Xb + Mc Xc). (iii) Calculate the polydispersity by using the answer in (ii) divided by answer in (i). (iv) If the molecular weight of repeat unit is 12, calculate the degree of polymerization by using the formula (Ma Xa + Mb Xb + Mc Xc) / (molecular weight of repeat unit).
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