ST5H.1- Suppose the mean ionic activity coefficient for the ions M2* and X1 of the ionic compound M1X2 is Y± = 0.791. If we could somehow magically know that y+ = 0.888 then what is y_? Type your answer..
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- Complete the balanced neutralization equation for the reaction below. Be sure to include the proper phases for all species within the reaction. П + Reset 0₂ C 1 2 3 4 HC₂H₂O₂(aq) + Sr(OH)₂(aq) 2 3 2 3 ■ U 00 Question 52 of 70 ■ O LO 5 6 + ☐6 ²+ 3+ 2+ 7 8 9 07 ☐8 H 4+ U 0 (s) (1) (g) (aq) Sr 口。 • x H₂O Delete Submit +A hollow square metal tube has a measurement of 2.42 in for each side and a weight of 53.71 g. The volume of empty space inside the tube was measured to be 0.93 in³. If the density of the metal used in constructing the tube is 11.81 g/mL, calculate the length of the tube in inches. Note: 1 in = 2.54 cm, 1 cm3 = 1 mL %3D Show your solutions in the next item Round your answer to 2 decimal places.KMnO4 and Na2C2O4 solutions were used in the reactions that took place in a back titration to determine the amount of H2O2 in a sample. Calculate the concentration of H2O2 in the sample (w / v) as% by making appropriate assumptions for the volumes and normality of all these solutions.
- Please in handwriting show how to solve for H+ ions -0.167=0.1400-0.0591/2 * log (0.108)/(H+)^2Standardization of a sodium hydroxide (NaOH) solution against potassium hydrogen phthalate (KHP) yielded the following results: Mass of KHP, g Volume of NaOH, mL 0.7987 38.29 0.8365 38.96 0.8104 38.51 0.8039 38.29 MM KHP: 204.22 MM NaOH: 40.00 a. Calculate the mean molar concentration of sodium hydroxide. (First, compute the concentration of NAOH in each titration then compute for its mean/average value.) b. Calculate the standard deviation. Please provide solutions for your answers. If you have used your calculator for calculating for mean and standard deviation, you can just provide the formula with the values indicated.Answer provided: 0.08432M Ag^+1 Please show your complete solution and write your answer clearly and readable. Thank you.
- Type your numerical answers in the box provided in three significant figures (e.g., 0.0123). Do not report your answer in scientific notation. No need for units. If Canvas drops your last zero digit/s, it is OKAY. (e.g. 1.20 to 1.2). Consider a solution of 0.0500 M HF (K= 7.20 x 10-4) for nos. 25 28. Determine the following: 25. equilibrium molar concentration of H3O%3= 26. equilibrium molar concentration of HF = 27. pOH = 28. %ionization %3D2- Suppose a 0.025 M aqueous solution of phosphoric acid (H3PO4) is prepared. Calculate the equilibrium molarity of HPO properties of phosphoric acid in the ALEKS Data resource. Round your answer to 2 significant digits. M 0 X S ? You'll find information on theHow does Sn3P4(s) break up in solution? Write subscripts and superscripts as normal numbers and do not include spaces in your answer. Do not include states like (aq) or "1" in front of single atoms or charges. For example, the answer for NiCl3is) would be Ni3++3CI- For example, the answer for Sb3(PO4)5(s) wOuld be 3Sb5++5PO43- Possibly relevant polyatomic ions: CN",OH", NO3", NO2', CIO3", CIO4", co,2", so,², Cro,²", Cr20,2", PO,, PO33- Screen Reader User: For example, the answer for NiCl3(s) would be Ni3++3Cl minus For example, the answer for Sb3(P O4)5(s) would be 3Sb5++5PO43 minus Possibly relevant polyatomic ions: 1 minus N O2 N03 1 minus CI O2 1 minus CI O, 1 minus minus CN minus 2 minus C2 2 minus s O, 2 minus Cr 0,2 minus, Crɔ O7 3 minus P O3 PO4 3 minus
- 3. The following molarities were calculated from replicate standardizations of a NaOH solution: 0.502 8, 0.502 9, 0.502 3, 0.503 1, 0.502 6, 0.502 7, 0.503 4, and 0.502 5. Assuming no systematic errors, within what range are you 90% certain that the true mean value of the molarity falls?Q1 A). Exactly 4.57g of BaCl2.2H₂O are dissolved in sufficient water to give 250 mL of solution. Calculate the formality? Ba= 137.3, Cl-35.5, H=-1, 0-16: knowing that B) Calculate the molarity of 28% NH3, specific gravity 0.898 solution? N=14, I1=1: knowing thatHomework Example Sample: NaC1+NaBr + inert excess AGNO, U = 1.000 g Ag: 107.87, Cl: 35.45 AgCl(s) + AgBr(s) = 0.5260 g Na: 22.99 Cl, treatment I AgCl(s) + AgCl(s) Br: 79.90 = 0.4260 g Nacl = ? % : NaBr = ? % Solution if NaCl=x g. NaBr =y g yx187.77 102.89 xx143.32 g of AgCl(s) = : g of AgBr(s) = 58.44 xx143.32 yx187.77 + xx100 = 0.5260 g % NaCl = 58.44 102.89 1.000 xx143.32 ух143.32 = 0.4260 g yx100 % NaBr =: 58.44 102.89 1.000