ENGINEERING MATHEMATICS - EXAMPLE 8.1 : A local utility burns coal having the following composition on a dry basis : Carbon (C) 83.05 %, hydrogen (H) 4.45 %, oxygen (O) 3.36 %, nitrogen (N) 1.08 %, sulfur (S) 0.7 % and ash 7.36 %. Calculate the ash free composition of the coal with reference to C, H, O, N and S.
ENGINEERING MATHEMATICS - ANSWER 8.1 : Total percentage without ash is 83.05 + 4.45 + 3.36 + 1.08 + 0.7 = 92.64 %. For ash free composition percentage calculations, C = 83.05 / 92.64 x 100 % = 89.65 %. H = 4.45 / 92.64 x 100 % = 4.80 %. O = 3.36 / 92.64 x 100 % = 3.63 %. N = 1.08 / 92.64 x 100 % = 1.17 %. S = 0.7 / 92.64 x 100 % = 0.76 %. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.6 : In N + 1 Rule in Quantum Chemistry, whenever a spin 1 / 2 nucleus is adjacent to N other nuclei, it is split into N + 1 distinct peaks. In 1 peak or singlet, there is only 1 magnitude. In 2 peaks or doublet, the ratio of magnitude of each peak is 1 : 1. In 3 peaks or triplet, the ratio of magnitude of each peak is 1 : 2 : 1. In 4 peaks or quartet, the ratio of magnitude of each peak is 1 : 3 : 3 : 1. In 5 peaks or quintet, the ratio of magnitude of each peak is 1 : 4 : 6 : 4 : 1. (a) By using binomial coefficients or Triangle of Pascal find the ratio of magnitude of each peak if 6 peaks exists. (b) How many adjacent nuclei are available in a spin 1 / 2 nucleus in such situation of 6 peaks?
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