Question 2: myo-Inositol is a growth factor in humans and micro-organisms, the structure of which is shown below. HO НО ОН ... ...ОН ОН ОН When draw in the LEAST stable chair conformation, how many -OH groups are placed in the EQUATORIAL position?

Consider the following reversible reaction at equilibrium: C6H12O6(aq) + 6 O2(g)      6 CO2(g) + 6 H2O(l). Given that this reaction is exothermic, if heat is added to the equilibrium system, how is the stress relieved?

Hydrogen peroxide, H₂O₂, has a density of 1.45 g/cm³ and decomposes spontaneously but slowly. The decomposition reaction will proceed at a faster rate if iodide ion, a catalyst, is added. The steps for the catalyzed reaction are: H₂O₂(aq) + (aq) → H₂O(l) + Ol-(aq) OO H₂O₂(aq) + Ol(aq) → H₂O(l) + O₂(g) +1-(aq) a. What is the net reaction? b. If 10.00 mL of H₂O₂ undergoes this reaction, what mass of oxygen is produced? c. If the reaction has an 85.5% yield, what mass of oxygen is actually produced?

16.

2NO(g) + O₂(g) → NO2(g) Conc. NO, mol/L Exp. 1 4.3 x 10-2 Exp. 2 4.3 x 10-2 Exp. 3 8.6 x 10-² Exp. 4 0.63 Conc. 02, mol/L 2.4 x 10-² 4.8 x 10-2 9.6 x 10-² 5.4 x 10-³ Initial Rate 8.0 x 10³ mol/(L.s) 1.6 x 102 mol/(L.s) 0.13 mol/(L.s) ? a What is the experimental rate law for the reaction above? (Use k for the rate constant.) Rate law = K[NO]²[0₂]✔ Co- The reaction is first order in O₂ because the rate doubled with a do second order in NO because the rate increased by a factor of 8 wh Rate = k[NO] ² [0₂] b What is the initial rate of the reaction in Experiment 4? Initial rate= mol/(L.s)

Consider the following equation:                        N2(g) + 2O2(g) + heat  <=>  2 NO2(g)  In what way will the equilibrium shift when the following stresses occur? The temperature is increased.  More N2 is added.  NO2 is collected and removed.  The pressure is increased.

Acetaldehyde (CH3CHO) is an important chemical both industrially and biologically.  For instance, it is a (somewhat toxic) intermediate in the body's metabolism of ethanol into acetic acid, and thus is possibly implicated in the "hungover" symptoms of someone who has had too much to drink the night before.  In aqueous solution, it establishes an equilibrium with a hydrated form, shown below. CH3CHO (aq)   +   H2O (l)   <-->  CH3CH(OH)2 (aq) You start with an aqueous sample, already at equilibrium, with the CH3CH(OH)2 (the hydrated form) at a concentration of 2.60 M.  You have no information about how much, if any, of the anhydrous form (CH3CHO) is initially in the flask.  If you add 2.0 M of CH3CHO to the reaction flask, and as the equilibrium is being restored the amount of CH3CH(OH)2 changes by 1.13 M, what is the final amount of CH3CHO?