Question

it so annoying to ask questions twice! just because a person can't answer properly! last person didn't solve it correct and I left a comment but unfortunately he/she was probably lazy to fix it! please make sure you are not lazy to help out! thank you

(1 point) Two chemicals A and B are combined to form a chemical C. The rate of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. Initially there are 58 grams of A and 40 grams of B, and for each gram of B, 0.8 grams of A is used. It has been observed that 24.5 grams of C is formed in 15 minutes. How much is formed in 20 minutes? What is the limiting amount of C after a long time? grams of Care formed in 20 minutes grams is the limiting amount of C after a long time

Answer 1

☆ x grams of chemical c at time t for each gram of B, 0.8 grams of so here total = 1 +0.8= 1.8 A used. ( 0.8 x grams of A) le grams of 8) so Amount of A and B remaining at time t is, (58 - 0.8 x) grams of A . (40-2) grams of B

☆ de estas (58- opa) (40 - ) de (56 - 05x) (60 - ( 104,4-0,5%) (42-x) dx = 0.8k (130,5 - x) (72-21) at 1812 dx = 0,8k dt (ac- 130.5) (x-72) - (- 130.5) (1.8) take ork k another constant 1.812 da : K (x - 130.5) (1-72). at da kat ( x 1 (x-72) = kat ( x- 130.) (2-72) use partial fraction, 1²/17 1 - 130.5 - 2/17 1dx = 31-72) take integration.

2 in (x-130.5) -2 ln(x-72) = kt to 117 117 | incx-120.5) – in 6-12)] - +C it ] * +6_ in City (kt+C) in/x-130.5 a-72 = 58.5kt + 58.5c use k 58.5k= 58.50 = constant = constant en /x-130,5) = kt tc 176-72 2-130.5 = ckttco C-72 use ě = c = constant

= c.ekt 26-130.5 2-72 X - 130.5 = cekt (x-72) x - 130.5 = caekt_72cekt a-cac ekt = 130:5-72cekt x (1-cekt) = 130,5-72cekt | OC = 130,5 - 72 CEC - - i-cekt initially, x0 =0 OE 1305. - 72 1 - ceº 0 = 130,5 – 72 ce o= 130.5 - 726 [c= 1.8125 put in ean 26 130, 5 - 72 (1.8125) et 1- 1.5125ekt

x = 130.5 - 130.5 okt 1-1,8125 ekt x = 130.5 (1- ekt) |- |.૨; ek divide by ekt be = 130.5(ékt - 1) / © é*t - 1.8125 24.5 grams of 15 minutes a formed in x(15) = 24.5 so 24.5 = 130.50 € RCS)-) e keis) - 1.6125 24.5 Cés -1.8125 ) =130.5e"" - 130.5 24,50 TSK - 130.50 156 = 44.4 - 130.5

-106 ek = -86.093 ~15k = 0.8122 - 15k = enc0.208) k = enco.208) -15 Ik=0.01356 put in ean X = 130,5 (e-0.01356€ -) 1.01366t -1.8125 a take t = 20 min. x = 130,5 (e 0.01386 (20) , -0.01386 (20) - 1.8125 De = 130 (0.7577 - V). 0.7577 -1.6125 x = 29.9679 grams of a formed in 20 minutes

☆ take t o in e x= 130.5 [ė 0.01386(0) - 17 -0.01386() - 1.5725 x=130,5 [e-a -1] € 0 -1.8125 2=130.5 (0-1) 0 -1.8725 2 = 130,5 1.8425 Tac=72 grams is limiting amount