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- A. 0.11MB. 0.36MC. 0.036MD. 0.0034MThe thermogram in the figure below shows the change in mass for a sample of calcium oxalate monohydrate, CaC2O4 × H2O. The original sample weighed 42.50 mg and was heated from room temperature to1500 °C at a rate of 5 °C min. The following changes in mass and corresponding temperature ranges were observed: • Loss of 4.03 mg from 100–250 °C• Loss of 5.72 mg from 400–500 °C• Loss of 8.41 mg from 700–850 °CDetermine the identities of the volatilization products and the solid residue at each step of the thermal decomposition.Types of Error; Propagation of Uncertainty from Random ErrorYou prepare an NH3 [FM 17.030 5 (±0.000 2)] solution by diluting 7.36 (±0.03) mL of a 28.00 (±0.05) wt% NH3 solution [density = 0.899 (±0.003) g/mL] to a final volume of 500.00 (±0.02) mL. Find the molarity of the NH3 solution along with its absolute uncertainty.
- The method described above is to be used for the analysis of ores that assay about 1.2% gold. What minimum sample mass should be taken if the relative error resulting from a 0.4-mg loss is not to exceed a. -0.1% b. -0.8%Ksp= 3.00 x 10^-17Volumetric Glassware and Balances In an experiment for calibration of a 5.00 mL. pipet, the following data was obtained for 5.00 ml. of the pipet at 15°C The correction <--0.04 mL The volume of 1 g of water at 15°C- 1.0021 mL The weight of water delivered by the pipet (g) is: a) 5.06 b) 4.92 c) 5.03 d) 4.95 e) 5.09
- Nn.127. Subject :- Chemistry3. A method of analysis yields masses of gold that are low by 0.4 mg. Calculate the percent relative error caused by this result if the mass of gold in the sample is a. 500 mg b. 250 mg c. 60 mg 4. The method described above is to be used for the analysis of ores that assay about 1.2% gold. What minimum sample mass should be taken if the relative error resulting from a 0.4-mg loss is not to exceed a. -0.1% b. -0.8%Round 12345.6 to 4 significant figures
- f the tolerance of a 50 mL Burette is \pm 0.05 mL, which of the following would not be acceptable readings to record? 9.3 mL 43.02 ml. 22.3 mL 0.65 mL 52.45 mL 30.07 mL 15.30 mL 36.55You have conducted an experiment where you obtained 22.4g ZnCl2. The accepted value is 22.7g. What is the error and relative (%) error?A series of standards were analyzed which gave the following results SOLUTION % Transmittance 0 mL of 0.100 mg/mL Hg + 2 mL tinchloride in a total volume of 75 mL 100 1.0 mL of 0.100 mg/mL Hg + 2 mL of tinchloride in a total volume of 75 mL 81.3 2.0 mL of 0.100 mg/mL Hg + 2 mL of tinchloride in a total volume of 75 mL 63.9 3.0 mL of 0.100 mg/mL Hg + 2 mL of tinchloride in a total volume of 75 mL 55 5.0 mL of 0.100 mg/mL Hg + 2 mL of tinchloride in a total volume of 75 mL 35.1 How would you make a Beer-Lambert Plot with this information?