The fact that there are two reactants, should make you wonder if the reaction is second order overall. However, in class, we presented data that showed that the natural log of the amount of light change over time was a straight line with a negative slope. This would tend to make you think that the reaction is first order. If we also knew that the reaction rate was dependent on luminol concentration, you would say that this reaction is:

Calculate the enthalpy of the following reaction: 4 b (s) + 3 o2 (g) → 2 b2o3 (s) given the following pertinent information: (a) b2o3 (s) + 3 h2o (g) → 3 o2 (g) + b2h6 (g), δhoa = +2035 kj (b) 2 b (s) + 3 h2 (g) → b2h6 (g), δhob = +36 kj (c) h2 (g) + latex: \frac{1}{2} 1 2 o2 (g) → h2o (l), δhoc = −285 kj (d) h2o (l) → h2o (g), δhod = +44 kj

Problem #3 (ch. 1, problem 15)the ideal gas law provides one way to estimate the pressure exerted by a gas on a container. the law isí‘ťí‘ť=푛푛푛푛푛푛푉푉mo re accurate estimates can be made with the van der waals equationí‘ťí‘ť=í‘›í‘›í‘›í‘›í‘›í‘›í‘ ‰í‘‰â’푛푛푟푟â’푞푞푛푛2í‘ ‰í‘‰2where the term nb is a correction for the volume of the molecules and the term an2/v2is a correction for molecular attractions. the values of a and b depend on the type of gas. the gas constant is r, the absolutetemperature is t, the gas volume is v, and the number of moles of gas molecules is indicated by n. if n = 1 mol of an ideal gas were confined to a volume of v = 22.41 l at a temperature of 0â°c (273.2k), it would exert a pressure of 1 atm. in these units, r = 0.0826.for chlorine gas (cl2), a = 6.49 and b = 0.0562. compare the pressure estimates given by the ideal gas law and the van der waals equation for 1 mol of cl2 in 22.41 l at 273.2 k. what is the main cause of the difference in the two pressure estimates, the molecular volume or the molecular attractions?

What must happen before a body cell can begin mitotic cell division