The reaction below has an equilibrium constant kp=2.2×106 at 298 k. 2cof2(g)⇌co2(g)+cf4(g) you may want to reference (pages 680 - 684) section 15.3 while completing this problem. part a calculate kp for the reaction below. 4cof2(g)⇌2co2(g)+2cf4(g)

Let the ed50 of a recreational drug be defined as the amount required for 50% of a test group to feel high or get a buzz. if the ed50 value of ethanol is 470 mg/kg body mass, what dose would a 70 kg party goer need to quickly consume in order to have a 50% chance of getting a buzz? 235 mg 470 mg 32,900 mg 35,000,000 mg