Acertain substance x decomposes. it is found that 50% of x remains after 100 minutes. how much x remains after 200 minutes if the reaction order with respect to x is a) zero order b) first order c) second order

Zero Order: t(1/2) = [A]o/(2*k), so k = [A]o/(2 * t(1/2)), where t(1/2) = 100 min. and [A]o=100% So incorporate condition. A= - .0005 x 200 mins+1 of introductory fixation approaches zero percent. 
Initially Order: Half-life condition k=.00693 remaining with 100 mins half life 
Coordinate condition lnx= - .00693x200mins + ln 1 squares with lnx= - 1.386. Take to the e to counteract ln so x=.25 or 25% 
Second Order Initial Concentration of 1, half-existence of 100 mins. K= .01. Coordinate condition 1/x = .01 x 200 +1. X= 33.3%

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na

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