A concentration, C (mol/L) varies with time, t (min) according to a valid empirical equation (Q-1) below, C = 3.00 e^-2.00t (Q-1) What are the units of 3.00 and 2.00? Calculate the concentration (C) in mol/m3 at t = 10 seconds.
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- A concentration, C (mol/L) varies with time, t (min) according to a valid empirical equation (Q-1) below,
- C = 3.00 e^-2.00t (Q-1)
- What are the units of 3.00 and 2.00?
- Calculate the concentration (C) in mol/m3 at t = 10 seconds.
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- Solve the required of the following and show the complete process to be able to gain the value of the correct answer. Given: Volume of Vinegar used (ml) = 25.00 mL Volume after dilution = 250.00 mL Volume of Aliquot used (ml) = 50.00 mL Volume NaOH used (ml) = 34.85 mL Concentration of NaOH = 0.09919475765 M Required: Calculate % (w/v) CH3COOH = ? P.S. Use weight in grams and volume of vinegar in mL, but correction based on aliquot should be made.Using dimensional analysis, find the mass of acetylalicylic acid in one tablet of aspirin. The grams of aspirin used in a titration is .325g before being dissolved in 10mL of water and then added to a base, NaOH, with a molarity of .01M and a volume of 10.1 mL. In Chem2.png, the actual question is boxed in red Video used for titration assignment (Not needed for question): https://www.youtube.com/watch?v=x9jgjdizTaYA stock solution of 0.225 +/- 0.003 M NaNO2 was transferred to a 100 mL volumetric flask (class A) and diluted to the mark. If 7 mL of the stock solution was transferred using one 5-mL and two 1-mL volumetric pipettes, what is the new concentration of the solution and the uncertainty? Please show how to calculate the uncertainty, that is the one part I don't understand how to do
- EX 2.Given-> Volume of Na+ = 500 ml Molarity of Na+= 0.0100M Molar mass of Na2CO3 = 105.99 gm/mole Millimole of Na+ = molarity × volume Number of millimole = 0.0100 × 500 = 5 millimole Na2CO3 ---> 2Na+ + CO32- Millimole of Na2CO3 = millimole of Na+/2 Millimole of Na2CO3 =5/2 = 2.5 millimole Mole of Na2CO3 = 2.5 × 10-3mole (1 mole = 10^3 millimole) Weight of Na2CO3 required = mole × molar mass = 2.5 × 10-3 × 105.99 =0.26 gm Hence, 0.26 gm Na2CO3 must dissolve in 500 ml of water.4. A fat sample with combination of acids contain standard hydrochloric acid for blank and sample with 8mL and 5mL respectively. The normality of the standard hydrochloric acid is 0.93N and the weight of the sample is 3 grams. Calculate the saponification value.Table of caffeine standards concentration . Sample Conc, ppm Std1 16 Std2 32 Std3 48 Std4 64 Std5 80 If the volume used to make 100 mL of std 1 is 2 uL what is the concentration in M used to make std calibration curve ? The standards are going to be used to build calibration curve to analyze caffeine in an energy drink. if 500 mL of the energy drink has target of 400 mg , how will you prepare the sample if you need 10 mL for the analysis ? Caffeine MM=194.19 g/mol.
- Given the following data, what is the molarity? Include the relative error Error is 1.0 mg or 1.0e-4I would need help with these questions. The method referenced above was followed by a student and she got the following results: Caffeine Std. Conc. (ppm) Absorbance 100 1.806 50 0.899 40 0.724 30 0.545 20 0.365 10 0.183 Further, she analyzed an unknown sample and she got the following results: Sample # Absorbance 1 0.398 1. Graphically find the concentration in ppm and then calculate in mol/L of Caffeine in Sample #1? 2. Which solvent was used to extract Caffeine and why? Why is it necessary to do extraction three times? 3. Cuvettes used in this experiment were made from which material and why? 4. At what wavelength the Absorbance was measured?Beaker 0.00200 M Fe(NO3)3, mL 0.00200 M NaSCN, mL total volume, mL 1 3.000 2.000 10.00 2 3.000 3.000 10.00 3 3.000 4.000 10.00 4 3.000 5.000 10.00 5 (blank) 3.000 0.000 10.00 In Solutions 1-4 you are adding successively larger volumes of 0.00200M SCN- to the Fe3+ solution and diluting to 10.00 ml. Calculate the final diluted molarity of SCN- in solution #1 Your answer should have 3 sig figs =
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