2.In a laboratory experiment you react 20.0 mL of 25% by mass C2H&N2with 8.000 NiCl2.6H20 to obtain the product Ni(NH2CH2CH½NH2j3Cl2 complex. TheC2H3N2 solution used in this experiment is 25.0% by mass, with a density (d) of0.950 g/mL.(MW NICI2.6H2O= 237.69 g/mole, MW C2HgN2=60.10 g/mole,MW tris(ethylenediamine)nickel(II)chloride = 309.89 g/mole)a. Write down the balanced equation of the synthesis.b. Determine the number of moles of each reactant that you use.c. Determine which reactant is the limiting reactant, giving your reasoning. Show allyour work to get full credit. (5 points)d. Calculate your percent yield if the dry product weighs 7.000 g.

The balanced chemical equation can be written as follows: NiH2O6Cl2 + 3C2H8N2 → NiC2H8N23Cl2 +…

2. In a laboratory experiment you react 20.0 mL of 25% by mass C2H&N2 with 8.000 NiCl2.6H2O to obtain the product Ni(NH2CH¿CHNH2y3Cl2 complex. The C2HạN2 solution used in this experiment is 25.0% by mass, with a density (d) of 0.950 g/mL. (MW NICI2.6H2O= 237.69 g/mole, MW C2H&N2=60.10 g/mole, MW tris(ethylenediamine)nickel(I)chloride = 309.89 g/mole) a. Write down the balanced equation of the synthesis. b. Determine the number of moles of each reactant that you use. c. Determine which reactant is the limiting reactant, giving your reasoning. Show all your work to get full credit. (5 points) d. Calculate your percent yield if the dry product weighs 7.000 g.

2. In a laboratory experiment you react 20.0 mL of 25% by mass C2H&N2 with 8.000 NiCl2.6H2O to obtain the product Ni(NH2CH¿CHNH2y3Cl2 complex. The C2HạN2 solution used in this experiment is 25.0% by mass, with a density (d) of 0.950 g/mL. (MW NICI2.6H2O= 237.69 g/mole, MW C2H&N2=60.10 g/mole, MW tris(ethylenediamine)nickel(I)chloride = 309.89 g/mole) a. Write down the balanced equation of the synthesis. b. Determine the number of moles of each reactant that you use. c. Determine which reactant is the limiting reactant, giving your reasoning. Show all your work to get full credit. (5 points) d. Calculate your percent yield if the dry product weighs 7.000 g.

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

[Review Topics] [References] In an aqueous chloride solution cobalt(II) exists in equilibrium with the complex ion CoCl42-. Co2+ (aq) is pink and CoCl42 (aq) is blue. At Low Temperature the pink color predominates. At High Temperature the blue color is strong. If we represent the equilibrium as: CoCl42 (aq) Co2+ (aq) + 4Cl(aq) We can conclude that: 1. This reaction is: A. Exothermic B. Endothermic C. Neutral D. More information is needed to answer this question. 2. When the temperature is increased the equilibrium constant, K: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question. Next - - - ) SONY DISPLAY WEB OFF One Touch Web Access Without Booting Up F11 F12 Num Lk Scr LK Prt Sc Sys Rg F10 Other Bookmarks THE 4:16 AM 4/30/2022 VGN-NW350F Pause Delete Break
(Q54) Nickel iodate, Ni(IO3)2 is being precipitated out of a 1.00 L solution. If the concentration of dissolved nickel is 0.292 M and the Ksp of nickel iodate is 4.71 x 10-5, what must the concentration of iodate be for precipitation to begin?
Explain the structure and magnetic property of [PtCl4]² complex by using CFT.
38 39 40 41 42 43 44 45 12- When dissolving iron in diluted hydrochloric acid, then adding NaOH Solution to the products. 1- The formed precipitate is ... (a) White gelatinous dissolve in sodium hydroxide solution. (b) Reddish brown precipitate dissolves in diluted hydrochloric acid (C) Pale green precipitate dissolves in diluted hydrochloric acid (d) Pale green precipitate dissolves in sodium hydroxide solution.
I need hand written solution only
The equilibrium constant for the following reaction is 1.0 x 102³. Cr³+ (aq) + H₂ EDTA² (aq) = CrEDTA¯(aq) + 2H+ (aq) 0₂C-CH₂- CH2–CO2 CH2CO, = EDTA+ O₂C-CH₂ Ethylenediaminetetraacetate EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt Na₂ H₂ EDTA, are used to treat heavy metal poisoning. Calculate [Cr³+] at equilibrium in a solution originally 0.0050 M in Cr³+ and 0.060 M in H₂EDTA² and buffered at 3+ pH 6.00. Assume that H₂ EDTA² form dominates at this pH. [Cr³+] = N- -CH₂-CH₂-N M
A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br- The liberated nickel ion required 11.70 mL of 0.002146 M EDTA. Calculate mmol Ni in the 50.00 mL aliquot. Calculate mmol Br- in the 50.00 mL aliquot. Calculate the percentage of NaBr (102.894) in the 1.000 g sample.
< 7:31 tetrachloronickelate(II) ([Ni(CI) 4]²-)? 1 What is the coordination number around the central metal atom in 4 7 +/- Question 24 of 28 2 5 8 3 6 9 O il 484 Tap here or pull up for additional resources Submit X с x 100
Kk.80.
For the aqueous [Hg14] complex K₁=6.76 × 10²⁹ at 25 °C. 2+ Suppose equal volumes of 0.0066M Hg(NO3), solution and 0.78M KI solution are mixed. Calculate the equilibrium molarity of aqueous Hg²+ ion. Round your answer to 2 significant digits. M x10 X ? Ś
The solubility constant Ksp of Ag2CrO4 is related to its solubility in water (s) as Ksp
A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br-Nickel ion required 11.70 mL of 0.002146 M EDTAWhat is the percentage of NaBr (102.894) in the 1.000 g sample?