Part b I ask that’s incorrect 2 of 6 blanks incorrectly Bse Homecom/course.html?courseld=17825142&OpenVellumHMAC=d853c82904958fb3726d3bc7b5c2554e#10001Imported From Sa...Q|MI|GQ|OB|A|GG|OQ|BookmarksBar ESPN y! Yahoo!HW15 // Adaptive Follow-Upestion for Group Work 20.142CAN2023.GEN2Cl₂(g) + H₂O₂(aq) → Cl(aq) + O₂(g) (acidic)Match the words in the left column to the appropriate blanks in the sentences on the right.000Submit Previous Answers Request AnswerF4X Incorrect; Try Again; 5 attempts remainingF5FacebookBalance non- H & O elements: H₂O₂ (aq) →O2(g) and Cl₂(g) → 2 Cl(aq)Separate: H₂O₂ (aq) →O2(g) and Cl₂(g) →Cl(aq)Cancel electrons: H₂O₂(aq) + Cl₂(g) + 2O₂(g) + 2 H+ (aq) + 2eAdd electrons: H₂O2(aq) →O2(g) + 2 H+ (aq) + 2e and Cl₂(g) +2 eAdd half-reactions: H₂O2(aq) + Cl2(g) →O2(g) + 2 H+ (aq) + 2 Cl(aq)Balance H with H: H₂O₂(aq) →O2(g) + 2 H+ (aq) and Cl₂(g) → 2 Cl(aq)MacBook AirF6Imported From Sa...57▶11CSReset Help+2 C1 (aq)Q ÛG Learn More2 Cl(aq)Hi, And

The given reaction is redox reaction . We need to balance this reaction in acidic medium. The…

Cl₂(g) + H₂O₂ (aq) → Cl(aq) + O2(g) (acidic) Match the words in the left column to the appropriate blanks in the sentences on the right. Balance non- H & O elements: H₂O₂ (aq) →O2(g) and Cl₂(g) → 2 Cl(aq) Separate: H₂O₂(aq) →O2(g) and Cl2(g) →Cl(aq) Cancel electrons: H₂O₂(aq) + Cl₂(g) + 2O₂(g) + 2 H (aq) + 2e Add electrons: H₂O₂ (aq) →O2(g) + 2 H+ (aq) + 2e and Cl₂(g) +2 e Add half-reactions: H₂O₂(aq) + Cl2(g) →O2(g) + 2 H+ (aq) + 2 Cl(aq) Balance H with H: H₂O₂(aq) →O2(g) + 2 H (aq) and Cl₂(g) → 2 Cl(aq) Reset +2 C1 (aq) 2 Cl(aq) Help

Cl₂(g) + H₂O₂ (aq) → Cl(aq) + O2(g) (acidic) Match the words in the left column to the appropriate blanks in the sentences on the right. Balance non- H & O elements: H₂O₂ (aq) →O2(g) and Cl₂(g) → 2 Cl(aq) Separate: H₂O₂(aq) →O2(g) and Cl2(g) →Cl(aq) Cancel electrons: H₂O₂(aq) + Cl₂(g) + 2O₂(g) + 2 H (aq) + 2e Add electrons: H₂O₂ (aq) →O2(g) + 2 H+ (aq) + 2e and Cl₂(g) +2 e Add half-reactions: H₂O₂(aq) + Cl2(g) →O2(g) + 2 H+ (aq) + 2 Cl(aq) Balance H with H: H₂O₂(aq) →O2(g) + 2 H (aq) and Cl₂(g) → 2 Cl(aq) Reset +2 C1 (aq) 2 Cl(aq) Help

Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Chapter33: Automated Methods Of Analysis

Section: Chapter Questions

Problem 33.6QAP

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Show all steps leading to the final answer po. Here’s a pdf file in accordance with the topic po: https://drive.google.com/file/d/1_FnDtXCrFKSol3RNWIG_9tNQ7IxgxD6t/view?usp=drivesdk
Please answer atleast 4 sub-parts. Thank you.
If 5 µM enzyme was used to obtain the data in the summary plot below, V-max = ________________ µM sec-1 and Km = ____________ µM for this enzyme (Enter numeric values to the nearest integer; Do NOT write units.)
from this pfd of nitrobenzene production show material balance at mixing -the conversion is 99% and production 1000 ton/day
Calculate the retardation factor (Rf) for dissolved Tetrachloroethene contamination given the bulk density of soil is 2.4 gm/cc, fraction organic content (foc) of soil is 0.005, and the Organic Carbon Partition Coefficient (Koc) is the number you look up in the table. HInt: Use Fetter Equation (Koc = Kd /foc to calculate Kd. Group of answer choices 1.4 to 2.7 l/gm 1.08 to 1.2 Cannot be calculated from given information. 3.25 to 4.5 1.05 to 1.2 l/gm
For Indirect Iodometric Analysis of Copper... ~0.0896g KIO3 necessary to consume 350mL of 0.1 M Na2S2O3, Na2S2O3 is stored in an amber glass bottle until ready for use. Primary Standard KIO3 has 2g of KI, 50mL of DI water, and 10 mL of 1.0M HCl is added then immediately titrated with Na2S2O3 until medium yellow or straw... then 5mL of starch indicator is added and titrated again until blue black color turns clear. Unknown CuO use 1.2G of Unknown, 20mL of HNO3 heated until sample dissolved, 25 mL of DI water added and boiled until clear light blue color, after cooling 1:1 NH3 added (~34.47 mL of NH4OH reagent) until permanent deep blue color amine complex, 2g of NH4HF2 added and swirled until dissolved, 3 g of KI is added then titrated immediately with Na2S2O3 until brown color of iodide is nearly gone (brown milk color), 2 g of KSCN and 3 mL of starch indicator is then added with titration continuing until disappearance of new blue black color. 1. Na2CO3 is often added to thiosulfate…
For Indirect Iodometric Analysis of Copper... ~0.0896g KIO3 necessary to consume 350mL of 0.1 M Na2S2O3, Na2S2O3 is stored in an amber glass bottle until ready for use. Primary Standard KIO3 has 2g of KI, 50mL of DI water, and 10 mL of 1.0M HCl is added then immediately titrated with Na2S2O3 until medium yellow or straw... then 5mL of starch indicator is added and titrated again until blue black color turns clear. Unknown CuO use 1.2G of Unknown, 20mL of HNO3 heated until sample dissolved, 25 mL of DI water added and boiled until clear light blue color, after cooling 1:1 NH3 added (~34.47 mL of NH4OH reagent) until permanent deep blue color amine complex, 2g of NH4HF2 added and swirled until dissolved, 3 g of KI is added then titrated immediately with Na2S2O3 until brown color of iodide is nearly gone (brown milk color), 2 g of KSCN and 3 mL of starch indicator is then added with titration continuing until disappearance of new blue black color. 4. Why is the starch indicator solution…
A chemist received different mixtures for analysis with the statement that they contained NaOH, NaHCO3, Na2CO3, or compatible mixtures of these substances, together with inert material. From the data given, identify the respective materials, and calculate the percentage of each component. 1.000g samples and 0.2500N HCl were used in all cases. (A) For Sample W: With PP, 24.32ml was used. A duplicate sample required 48.64ml with MO. (B). For Sample X: The addition of PP caused no color change. With MO, 38.47ml of the acid was required. (C). For Sample Y: To cause a color change in the cold with PP, 15.29ml of the acid was necessary, and an additional 33.19ml was required for complete neutralization. (D) For Sample Z: The sample was titrated with acid until the pink of PP disappeared; this process required 39.96ml. On adding an excess of the acid, boiling, and titrating back with alkali, it was found that the alkali was exactly equivalent to the excess acid added.
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Here's the link https://www.youtube.com/watch?v=wY3ey4E-FRA