Example Sample: NaCl+NaBr + inert = 1.000 g excess AGNO; U Ag: 107.87, Cl: 35.45 AgCl(s) + AgBr(s) = 0.5260 g Na: 22.99 Cl, treatment U AgCl(s) + AgCl(s) Br: 79.90 = 0.4260 g NaCl = ? % : NaBr= ? % %3D
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- Q5: If the Ksp of Ca3N2 is 6.06 x 10-3, how many moles of calcium nitride (s) could you dissolve in 1.589 L of solvent (assume that calcium nitride (s) dissolves in water to form Ca2+(aq) and N3- (aq))? Enter your answer with at least 3 sig figs. Please provide only typed answer solution no handwritten solution needed allowedWhat mass of sucrose, C12H22O11, must be dissolved per liter of water (d= 0.998 g/ml) to obtain asolution with 2.75 mole percent C12H22O11?Ksp for BaCrO4 is 1.17x10^-10
- Classify pure urea, H2NCONH2 fertilizer as NPK. Write your numerical answer in the order N, followed by P, followed by K, i.e., without using colon (:). E.g., if the answer is 10N, 25P, 33K, then write your answer as 102533.In a solution containing 25ml (NH4)2S2O8 with a total volume = 100 ml =0.1 liters , the solution composition are shown below.Note; the solution require 25ml of 0.2M S2O8^-2.SOLUTION Kl KNO3 EDTA Na2S2O3 STARCH50ml 23ml 1 drop 1 ml 10 dropsExperiment; initial (S2O8-2) =0.05M ; INITIAL (l-)=0.10 Mml of S2O3^-2 added Time in minutes and seconds for color change Cumulative time in seconds Total moles of S2O8^-2 consumed1 1:43 2.0 x 10^-42 1:20 4.0 x 10^-43 1:13 6.0 x 10 ^-44 1.17 8.0 x 10 ^-45 1.24 10 x 10 ^-41a) Find the cumulative time in secondsb) for the runs ,plot mole of S2O8^2- REACTION vs. time in seconds2) Draw a straight line through ( origin ) the points and calculate the slope3) divide the value of the slope by the total volume of ( 0.1 L) to get the rate in units of M/s4) find the rate5 )calculate x and y and k1) a)What is affinity chromatography and what is it used for? b) A drug called Bamlanivimab is used to treat infections with SARS-CoV-2 (Covid-19).This drug was developed by isolating antibodies from the serum of a patient who recovered from Covid-19.What is possibly on the stationary phase in a column used to isolate antibodies to SARS-CoV-2? How are the antibodies eluted off the column to isolate them for drug development?
- Determine the possible ion/cation present CATION CONFIRMATORY TEST OBSERVATION ADDED REAGENT(S) OBSERVATIONS Excess NH3 NVR KSCN NVR K4Fe(CN)6 NVR K3Fe(CN)6 NVR Na2C2O4 NVR NaOH (in evaporating dish) Litmus paper: red to blue ANION CONFIRMATORY TEST OBSERVATION ADDED REAGENT(S) OBSERVATIONS HNO3, Fe(NO3)3 Blood red aqueous layer; colorless toluene layer HNO3, KMnO4 Peach aqueous layer; colorless toluene layer (to the ppt) CH3COOH White ppt does not dissolve MgCl2 NVR 6M H2SO4, FeSO4, 18M H2SO4 NVR POSSIBLE ION(S) PRESENT? Select one or more: NH4+ CO32- Fe3+ SO42- Cu2+ NO3- Ca2+ I- Br- PO43- Zn2+ SCN-A 500.00 mg vitamin C (MW176.12g/mol) tablet was ground, acidified, and dissolved in H2O to make a 250.0 mL solution. A 50.00 mL aliquot containing vitamin C, KI and starch was analyzed and titrated with 12.31 mL of 0.01042 M KIO3. What is the % (w/w) vitamin C in the tablet? KIO3 + 5 KI + 6 H+ = 3 I2 + 6 K+ + 3 H2O C6H8O6 + I2 = C6H6O6 + 2I- + 2H+ A 67.77 % B 33.89 % C 22.59 % D 3.39 %Aspartic acid is a polyprotic system whose forms can be written as H3A+, H2A, HA-, and A2-. Its Ka values are: Ka1 = 1.02 x 10-2 Ka2 = 1.26 x 10-4 Ka3 = 9.95 x 10-11 How many milliliters of 0.30 M NaOH should be added to 100 mL of a solution containing 8.0 g of the neutral, zwitterionic form of aspartic acid (M.W. = 133.10 g/mol) to bring the pH to 10.3? Find the answer to the nearest mL.
- Aspartic acid is a polyprotic system whose forms can be written as H3A+, H2A, HA-, and A2-. Its Ka values are: Ka1 = 1.02 x 10-2 Ka2 = 1.26 x 10-4 Ka3 = 9.95 x 10-11 How many milliliters of 0.30 M NaOH should be added to 100 mL of a solution containing 8.0 g of the neutral, zwitterionic form of aspartic acid (M.W. = 133.10 g/mol) to bring the pH to 10.3? Find the answer to the nearest mL. The answer is between 300 and 400. Retain 4 d.p. in your calculations and don't round until you calculate the volume of NaOH required at the very end.You are trying to come up with a drug to inhibit the activity of an enzyme thought to have a role in liver disease. In the laboratory the enzyme was shown to have a Km of 1.0 x 10-6 M and Vmax of 0.1 micromoles/min.mg measured at room temperature. You developed a competitive inhibitor. In the presence of 5.0 x 10-5 M inhibitor, the apparent Km of the enzyme was found to be 1.5 x 10-5 M. What is the Ki of the inhibitor?5.00 mL of stock solution is diluted to 25.00 mL, producing solution ALPHA. 10.00 mL of solution ALPHA is diluted to 25.00 mL, resulting in solution BETA. 10.00 mL of solution BETA is then diluted to 25.00 mL, producing solution GAMMA. dilution factor for ALPHA from stock solution = 0.167 dilution factor for BETA from ALPHA solution = 0.0476 part c and d?