harsha

Find DHo for BaCO3(s) --> BaO(s) + CO2(g) given 2 Ba(s) + O2(g) --> 2 BaO(s) DHo = -1107.0 kJ Ba(s) + CO2(g) + 1/2 O2(g) --> BaCO3(s) DHo = -822.5 kJ 1. -1929.5 kJ 2. 537.0 kJ 3. 269.0 kJ 4. -284.5 kJ 5. -1376.0 kJ i got 537.0 kJ....is that right :S HELP...

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Use the upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width). (Round your answers to three decimal places.) y = sqrt of (1−x^2)

i think... work done=0.5*m*v^2 f*d=0.5*m*v^2 e*q*d=0.5*m*v^2 (v/d)*q*d=0.5*m*v^2 velocity=sqrt((2*100*3*e)/(2*Mn+2*Mp)) r=((2*Mn+2*Mp)*velocity)/(b*3*e))

thanks bro