Fill in the empty cells. Isotopic Symbol Element atomic Symbol number mass # of # of # of number protons neutrons electrons Format: X-# e.g. C-12 (1) Cs 135 (2) 83 209 (3) 82 56 (4) 120 50 (5) 78
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- The accompanying data (1.00-cm cells) were obtained for the spectrophotometric titration 10.00 mL of Pd(II) with 2.44 10-4 M Nitroso R(O. W Rollins and M. M. Oldham, Anal. chem .,1971, 43, 262, DOI: 10.1021/ac60297a026). Calculate the concentration of the Pd(II) solution, given that the ligand-to-cation ratio in the colored product is 2:1R 400m N CO₂tBu OH CO₂Me Ph LDA THF, -78 °C silica gel 6 days room temperature R MeO₂C NCO₂Bu CHO CO,Me I + H₂OWhat is species of a mineral with (Fe0.7Ni0.3)2SiO4 composition?
- 47. The maximum safe [Pb2+]=4.8x10-8M. Maintaining 1.0x10-4M levels for one of the anions below could ensure that this level is achieved by (Ksp of PbF2=3.6x10-8, PbCl2=1.7x10-5, PbCO3=7.4x10-14, and PbSO4=6.3x10-7) a. fluoridation b. chlorination c. adding sulfates d. carbonationOne common way to determine Phosphorous in urine is to treat the sample after removing the protein, with molybdenum (VI) and then reduce the 12-molybdophosphate with ascorbic acid to give an intense blue color species called molybdenum blue. The Absorbance of molybdenum blue can be measured at 650 nm. A patient produced 1122 mL of urine in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00 mL. A calibration plot was prepared by treating 1.00 mL aliquots of Phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm, and the following results were obtained: Solution absorbace at 650 nm 1.00 ppm P 2.00 ppm P 3.00 ppm P 4.00 ppm P Urine sample 0.230 0.430 0.630 0.840 0.518 (1) Construct a plot of the calibration curve and find the slope, intercept and the number of ppm of P in the sample. (ii) What mass in grams of phosphorous was…what is the thore tical yield of vanadium that can be produced y the reaction of 40.09 of V 205 with 40.0s of calcium based on the followin) chemiAl reacton V 205 CS) +5 Cal ) 2Va) +S CaD cS) 6-) 1.23 bi) 22ug c)stg d.) 40.05 e) 20:35
- According to the table below, Zn- has a slightly larger mobility than Na". How can you explain that? Table 19B.2* lonic mobilities in water at 298 K, u/(10-m?s-V-1) u/(10-m's- V-) u/(10- m²s"!V-) H+ 36.23 OH- 20.64 Na* 5.19 CF 7.91 K+ 7.62 Br 8.09 Zn2+ 5.47 8.29rd > My courses > (1_101232) (2) giac ê elas > 25 April-1 May > MidExam-28-4-2021 pard Site home Calendar Badges All courses estion 6 Which is the correct name for the complex [FeF4(OH2)2]¯ ? yet answered a. diaquatetrafluoroferrate(I) ion rked out of 2.00 Flag question O b. diaquatetrafluoroferrate(III) ion O c. diaquatetrafluoroiron(III) ion O d. diaquatetrafluoroiron(I) ion PREVIOUS PAGEGiven A, B, and C… [A] U(VI) as uraninite; UO2 (where Fe2+= reductant; Fe(OH)3 ferrihydrite= product): 2Fe2+ + UO22+ +3H2O + H+ ---- > 2Fe(OH)3 + U4+ +2H2O [B] U(VI) as uraninite; UO2 (where Mn2+= reductant; MnO2 pyrolusite= product): 2H2O + UO22+ + Mn2+ ---- > UO2 + B - MnO2 + 4H+ [C] U(VI) as as uraninite; UO2 (where HS-= reductant; S0= product): UO22+ + Hs- ---- > UO2 + S + H+ QUESTION: Use thermodynamic calculations [use redox potential (Eh)] to predict which of the three reductants below is most favorable at pH 3 ? (1) Fe2+ (2) Mn2+ (3) HS-
- A Part B Choose the structure with highlighted donor atoms. OH₂N H2 =0: :0: O H₂NT H₂N H₂ H₂ 0: OH₂N H₂ !ö: !ö: :0: Submit Request Answer Part C Give the coordination number of the metal in the complex. Express your answer as an integer. ΤΟ ΑΣΦ -> 0 i ? MacBook AirFor a solution in which µ = 6.5 × 10-2, calculate K'sp for a. AgSCN (Ksp (AgSCN) = 1.1 x 10-¹2, KSP b. PbI2 = = a Ag = -9 ², αpb²+ = 0.45, al- = 0.3) (Ksp (PbI₂) = 7.9 × 10¯ KSP c. La(IO3)3 -11 (Ksp (La(103)3) = 1.0 × 10¯ KSP = = 0.25, ascN- = 0.35) , αLa³+ 3+ =0.9, α103- - = 0.35) =From the Hill Plot below, the KD of the first binding event for the receptor-ligand system under study is: 10 8 6 4 2 0 -2 4 6 8 10 (0=1) 601 log -4- -6 -6 -4 -2 0 2