Sarah, do you know how to calculate the amount of NaBr formed (this is a limiting reagent problem). If you do,that calculation gives you the amount of NaBr under 100% yield.
% yield = 100(actual yield/1--% yield)
In a reaction involving the bromination of acetone, the following initial concentraions were present in the reaction mixture. acetone (0.8mol/L), H+ (0.2 mol/L), Br2 (0.004 mol/L) At 25 degrees C, it took 240s before the colour of
The reaction is Br2 (g) <---> 2Br (g). It occurs at T = 1600 C. When 1.05 mol Br2 are placed in a 2 Liter flask, 2.50% of Br2 undergoes dissociation. Calculate the Kp for the reaction. Okay so i did the ICE table for this,
For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants? 2Na(s)+Br2(g) ---->2NaBr(s) 1. 2mol Na, 2mol Br2 2. 1.8mol Na, 1.4mol Br2 3. 2.5mol Na, 1mol Br2 4. 12.6mol Na, 6.9 mol
The red-ox reaction is MgSO4+2K yields K2 SO4+Mg. I thought the oxidation reaction is 2 k yields K2(subscript) + e^-2. I said the reduction half reaction was Mg+e- yields Mg. Are these correct? Thanks for checking.
A 0.960mol quantity of Br2 is added to a 1.00 L reaction vessel that contains 1.23mol of H2 gas at 1000 K. What are the partial pressures of H2, Br2, and HBr at equilibrium? At 1000 K, Kp=2.1×106 and ΔH∘ = -101.7 kJ
For the reactions that occurred spontaneously in the individual tables, balance these reactions assuming they are in an acidic solution. Reaction 1: Cu I^- Ag Br^- Reaction 2: Zn Pb Cu Ag Reaction 3: Cl2 Br2 I2 Reaction 1 I 2(aq)