Answer / kang chuen tat (malaysia - pen

Answer 48 - (a) V = 0.3349 x 0.3349 x 0.3349 = 0.0376 cubic nanometers. (b) Mass per unit cell = 2 x 92.9 (g / mol) [ 1 mol / (6.02 x 10 ^ 23 atoms) ] = 3.086 x 10 ^ (-22) g = m. V = 0.0376 x [ 10 ^ (-7) ] ^ 3 = 0.0376 x 10 ^ (-21) cubic cm. Then density is m / V = 8.207 g / cubic cm. 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.

What are the various utilities of the process plant?

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What are the problems that are associated with the expansion joints?

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I want previous year question paper of ONGC for chemical engg.

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Question 72 - (a) According to United States Department of Agriculture (USDA) (http://ndb.nal.usda.gov/ndb/search/list, accessed 12 August 2016), 100 g of potatoes generate 77 kcal of energy. For raw tomatoes, 111 g have 18 kcal of energy. Question : How much energy will one gain if 150 g of heated potatoes are eaten with 200 g of raw tomatoes? (b) If 1 Calorie = 1 food Calorie = 1 kilocalorie and 1000 calories = 1 food Calorie, then how many Calories are there in 9600 calories? (c) According to a food package of potato chips, 210 Calories are produced per serving size of 34 g. In actual experiment of food calorimetry lab, 1.75 g of potato chips, when burnt, will produce 9.6 Calories. For each serving size of potato chip, find the difference of Calories between the actual experimental value and the value stated on the food package. (d) The specific heat of water is c = 1 cal / (g.K) where cal is calory, g is gram and K is Kelvin. Then what is the temperature rise of water, in degree Celsius, when 150 g of water is heated by 9600 calories of burning food?

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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.2 : (a) In the pricing of a coupon bond, the formula is : P = c / (1 + r) + (c + B) / [ (1 + r) (1 + r) ] for 2 years to maturity, where c = annual coupon payment (in dollars, not a percent), B = par value, P = purchase price. Five years ago someone bought a 20 year coupon bond and would like to get rid of it now : A coupon rate of 7 %, it matures in exactly 2 years, par value is $1000, current interest rate is 5 %. (i) Find the value of c. (ii) Find the value of r or interest rate. (iii) Find the value of P. (iv) Guess the formula for P when the maturity period is 3 years, if such formula for 1 year duration is P = (c + B) / (1 + r). (b) In lemma of Ito on a Forward, recall that a forward contract is priced at : ln F = ln S + rT. Find the value of F in 5 decimal points when S = $100, r = interest rate = 0.05 / year, T = duration = 1 year.

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Question 67 – In order to predict the wastewater production, the population number has to be understood. The population data is : 72000 (for year 1961 or P-1961), 85000 (for year 1971 or P-1971), 110500 (for year 1981 or P-1981). (a) Find the average population increase, or [ (P-1981 - P-1971) + (P-1971 - P-1961) ] / 2. (b) Find the average percentage population increase, or [ (P-1981 - P-1971) / P-1971 + (P-1971 - P-1961) / P-1961 ] / (2) X 100. (c) Find the incremental increase or P-1981 - 2 (P-1971) + P-1961. (d) Let Po = P-1981. After 2 decades or n = 2, the population is P-2001. By using arithmetical increase method, find P-2001 = Po + n (Answer for a). (e) By using incremental increase method, find P-2001 = (Answer of d) + n (n + 1) (Answer of c) / 2. (f) By using geometrical increase method, find P-2001 = Po [ 1 + (Answer of b) / 100 ] ^ n where ^ is power sign, or 1 ^ 2 = 1 x 1 = 1. (g) If the actual P-2001 = 184000, which method of estimation is more accurate, based on your answer in (d), (e) and (f)?

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HEAT TRANSFER - EXAMPLE 5.3 : In a cylinder with a hollow, let a is outside radius and b is the inside radius. In a steady state temperature distribution with no heat generation, the differential equation is (d / dr) (r dT / dr) = 0 where r is for radius and T is for temperature. (a) Integrate the heat equation above into T(r) in term of r. (b) At r = a, T = c; at r = b, T = d. Find the heat equation of T(r) in term of r, a, b, c, d.

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CHEMICAL MATERIAL BALANCE – EXAMPLE 2.2 : Three hundred gallons of a mixture containing 75.0 wt % ethanol and 25 wt % water (mixture specific gravity = 0.877) and a quantity of a 40.0 wt % ethanol - 60 wt % water mixture (specific gravity = 0.952) are blended to produce a mixture containing 60.0 wt % ethanol. The specific gravity of a substance is the ratio of density of a substance compared to the density of water. The symbol of weight percent is wt %. (a) Estimate the specific gravity of the 60 % mixture by assuming that y = mx c where y is wt % ethanol, x is mixture specific gravity. Values for m and c are constants. (b) Determine the required volume of the 40 % mixture.

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Tell us about a time when you failed to meet a deadline. What were the repercussions?

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