A 165.0 mL solution of 2.925 M strontium nitrate is mixed with 220.0 mL of a 3.221 M sodium fluoride solution.Calculate the mass of the resulting strontium fluoride precipitate.mass:gAssuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter aO for the concentration.[Na*] =M[NO,] =[S2*] =M[F] =M10:16 PM4)ENG3/26/2021lypBANG &OLUf12ins"prt scdeletehomeendpg up

A numerical problem based on molarity, which is to be accomplished.

A 165.0 mL solution of 2.925 M strontium nitrate is mixed with 220.0 mL of a 3.221 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate. mass: Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a O for the concentration. [Na*] = M [NO,] = M [S2*] = %3D M [F] = M

A 165.0 mL solution of 2.925 M strontium nitrate is mixed with 220.0 mL of a 3.221 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate. mass: Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a O for the concentration. [Na] = M [NO,] = M [S2] = %3D M [F] = M

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Gold has compounds containing gold(1) lon or gold(III) lon. A compound of gold and chlorine was treated with a solution of silver nitrate, AgNO3, to convert the chloride lon in the compound to a precipitate of AgCl. A 1.791 mg sample of the gold compound gave 1.105 mg AgCl. Calculate the percentage of the chlorine in the gold compound. Percentage =
A gravimetrical analysis is done with an unknown sample containing AgNO3. A 1.5020-g solid mixture is dissolved in water and treated with an excess of NaCl solutions, resulting in the precipitation of 0.6168 g of AgCl. What is the mass of AgNO3 in the mixture?
A 1.50 g sample of KCl is dissolved in 36.0 mL of water. The resulting solution is then added to 32.0 mL of a 0.470 M CaCl2 (aq) solution. Assuming that the volumes are additive, calculate the concentrations of each ion present in the final solution. Concentration of K* ions = M Concentration of Ca2+ ions = M Concentration of Cl ions = M
Suppose 31.0 mL of 0.510 M CoCl2 solution is added to 72.0 mL of 0.180 M NiCl2 solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing. Assume that the volumes are additive. Concentration of Co2+ = mol/L Concentration of Ni2+= mol/L Concentration of Cl- = mol/L
A student mixes 100.0 mL of 0.500 M AgNO3 with 100.0 mL of 0.500 M CaCl2. Write the balanced molecular equation for the reaction. Write the net ionic equation for the reaction. How many grams of precipitate will form? What is the concentration of Ag+, NO3‒, Ca2+, and Cl‒ in the final solution (assume volumes are additive).
In the laboratory you are given the task of separating Ca2+ and Fe2+ ions in aqueous solution. For each reagent listed below indicate if it can be used to separate the ions. Type "Y" for yes or "N" for no. If the reagent CAN be used to separate the lons, give the formula of the precipitate. If it cannot, type "No" Y or N 1. 12. 3 Reagent Formula of Precipitate if YES ка K₂SO4 Na CO
A 4.277-g sample of Al2(SO4)a was dissolved in distilled water to make a total volume of 250.0 mL of solution A. Then an aliquot of 25.00 mL of solution A was diluted 10 times with distilled water to give solution B. Determine the molar concentration of the sulfate ions in solution B.
Barium can be analyzed by precipitating it as BaSO4BaSO4 and weighing the precipitate. When a 0.265 gg sample of a barium compound was treated with excess H2SO4H2SO4 8.50×10−2 gg of BaSO4BaSO4 formed. What is the percentage of barium in the compound?
4. A solution of 75 mL NazCO3 is neutralized with 67.2 mL of 0.1065 M HCI. In the another analysis, 50 mL NazCO3 is added to a solution of 25 mL Ba(NO3)2, and after the barium carbonate was precipitated, the excess of NazCO3 is neutralized with 14.8 mL of the same acid. According to this, what is concentration of barium nitrate (261.337 g/mol) in molarity and g/L?
One method of determining the concentration of hydrogen peroxide (H₂O₂) in a solution is through titration with iodide ion. The unbalanced net ionic equation for this reaction is H₂O₂ +1+H* → 1₂ + H₂O A 50.00 mL sample of a hydrogen peroxide solution is found to react completely with 37.12 mL of a 0.1500 M KI solution. What is the concentration of hydrogen peroxide in the sample? O a. 0.2227 M O b. 0.4041 M O c. 0.1010 M O d. 5.568 x 10-² M 17.5 mL of a 0.1050 M Na₂CO3 solution is added to 46.0 mL of 0.1250 M NaCl. What is the concentration of sodium ion in the final solution? O a. 0.119 M O b. 0.148 M O c. 0.539 M O d. 0.205 M
You mix 29.0 mL of 0.312 M FeCl3 with 38.5 mL of 0.446 M NaOH. What mass of Fe(OH)3 (in grams) will precipitate from this reaction mixture? Mass= g One of the reactants (FeCl3 or NaOH) is present in a stoichiometric excess. What is the molar concentration of the excess reactant remaining in solution after Fe(OH)3 has been precipitated? Assume the volumes are additive. Concentration = M
d) A 38.5mL sample of 0.875M AgNO, solution is mixed with 51.0mL sample of 0.744M NaCl solution. The AgCl precipitated has a mass of 2.06g. Calculate the limiting reactant, the theoretical yield, and percent yield. AGNO3(aq) + NaClaq) → AGCI6) + NaNO3(aq)