Redox Titration Problem 1 What is the concentration of the chromium (1) ions in a solution obtained from the analysis of

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899603
Joined: Mon Aug 02, 2021 8:13 am

Redox Titration Problem 1 What is the concentration of the chromium (1) ions in a solution obtained from the analysis of

Post by answerhappygod »

Redox Titration Problem 1 What Is The Concentration Of The Chromium 1 Ions In A Solution Obtained From The Analysis Of 1
Redox Titration Problem 1 What Is The Concentration Of The Chromium 1 Ions In A Solution Obtained From The Analysis Of 1 (384.73 KiB) Viewed 10 times
Redox Titration Problem 1 What is the concentration of the chromium (1) ions in a solution obtained from the analysis of a stainless steel fork? (0.0254 mol/L) Experimental Design A 0.0114 mol/L potassium permanganate solution is titrated to oxidize 25.0 mL of chromium (II) ions to chromium (III) ions in an acidic solution. Evidence Volumes of Potassium Permanganate Solution Required Trial 1 2 3 Final Buret Reading (mL). 14.2 25.4 36.5 Initial Buret Reading (mL) 2.7 14.2 25.4 4 47.6 36.5 Analysis Problem 2 What is the concentration of the nickel (II) ions in a solution obtained in the analysis of nichrome wire? (0.0644 mol/L) Experimental Design A 0.0105 mol/L potassium permanganate solution is titrated to oxidize 10.0 mL of nickel(II) ions to nickel (III) ions in an acidic solution. Evidence Volumes of Potassium Permanganate Solution Required Trial 1 2 3 Final Buret Reading (mL). 15.8 28.1 40.4 Initial Buret Reading (mL) 2.7 15.8 28.1 4 42.9 30.7 Analysis
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply