The balanced equation for the reaction between NaOH and aspirin is as follows;
NaOH + C₉H₈O₄ --> C₉H₇O₄Na + H₂O
stoichiometry of NaOH to C₉H₈O₄ is 1:1
The number of NaOH moles reacted - 0.1002 M / 1000 mL/L x 10.00 mL
Number of NaOH moles - 0.001002 mol
Therefore number of moles of aspirin - 0.001002 mol
Mass of aspirin reacted - 0.001002 mol x 180.2 g/mol = 0.18 g
However the mass of the aspirin sample is 0.132 g but 0.18 g of aspirin has reacted, therefore this question is not correct.
Electrochemical cell representation for above reaction is,
Br-/Br2//I2/I-
Reaction at Anode: Br2 + 2e- → 2Br- (1)
Reaction at Cathode: 2I- → I2 + 2e- (2)
Standard reduction potential for Reaction 1 = Ered(anode) = 1.066 v
Standard reduction potential for Reaction 2 = Ered(cathode) = 0.535 v
Eo cell = Ered(cathode) - Ered(anode)
= 0.535 - 1.066
= -0.531v
Now, we know that ΔGo = -nF (Eo cell) ..............(3)
Also, ΔGo = RTln(K) ..........(4)
Equation 3 and 4 we get,
ln (K) = nF (Eo cell) / RT
= 2 X 96500 X (-0.531)/ (8.314 X 298)
∴ K = 1.085 X 10^-18.
the emission of ionizing radiation or particles caused by the spontaneous disintegration of atomic nuclei.
A because you can prove that with the graph you are given in the wuestion