According to dilution law, C1 * V1 = C2 * V2
For oxalic acid
C1 = 0.639 M , V1 = 5.00 mL , V2 = 12.00 mL
C2 = C1 * (V1 / V2)
C2 = 0.639 M * (5.00 mL / 12.00 mL)
C2 = 0.266 M
Adjusted molarity of oxalic acid = 0.266 M
For KMnO4
C1 = 0.150 M , V1 = 1.00 mL , V2 = 12.00 mL
C2 = C1 * (V1 / V2)
C2 = 0.150 M * (1.00 mL / 12.00 mL)
C2 = 0.0125 M
Adjusted molarity of KMnO4 = 0.0125 M
Average time = (time taken for trial 1 + time taken for trial 2 + time taken for trial 3) / 3
Average time = (407 s + 360 s + 367 s) / 3
Average time = 378 seconds
Rate of reaction = (Adjusted molarity KMnO4 / average time)
Rate of reaction = (0.0125 M / 378 s)
Rate of reaction = 3.31 x 10-5 M/s
The study of kinetics ( need to show the calculations) Final concentro ion THE STUDY OF...
need help with the Molarity of both parts
EXPERIMENT 8: AN OXIDATION-REDUCTION TITRATION DATA SHEET 0.2M 10 m2 Concentration of original KMnO solution Volume of Original KMnO4 solution used Molarity of KNOWN HOCg04 D.ISOM Standardization of KMnO solution. Sample 3 Sample 1 Volume H2C204 11.00ML Initial buret reading (KMnO4) .00 mc Final buret reading (KMnO4) 12.00ML Volume KMnO4 12.00ml Molarity KMnO4 Average (KMnO4) Sample 2 4:00ML 12.00mL 13.5ML . SOL Determination of Unknown Solution Sample 3 Sample 1 4ml Volume...
Can you please show how to do the calculations I sent
for run 2.
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