NaHCO, when treated with acids it may evolve CO. Acids which are stronger & comparable 30. For the given compounds NH, evolve CO (b) (c) (d) (B)ad>b>c (C)b>c>d>a (D)a>d>c>b (a) (A)c>b>da

(NH3] [H3o GREESHMA A.S ealculatee the PH Volue of 0.10M Na HZP04 1. Solu tion · kat (Ha Po4) = 4·52X10-3 kaz CHs PO4) = 7•52X10-3 > ka 3 (Ha Po4) = 4.8 A10-13 %3D kaz CH3 Po4)- 6•23110

Give the major organic products from the following reactions

The following equation describes the equilibrium that exists when the weak acid propionic acid (C2H5 CO,H) dissolves in water. (In acting as an acid, the -CO2H group supplies H+ to form H30*.) C2H; CO,H(aq) +H2O(l) 2 H,0+ (aq) + C2 H; CO2-(aq) a Which of the four species are Bronsted-Lowry acids? (Select all that apply.) O H20 O C2H; CO2- O H3o+ O C2H; CO,H b Which of the four species are Brønsted-Lowry bases? (Select all that apply.) O H2O O C2H;CO2 O H30+ O C;H; CO,H

Hello there, you solved first three sub-parts for me. now ı need the last three sub-parts thank you

PLUS G The value of Kb for x M Mathway Calculu X G What is the formol X G hcn weakest acid x G n/takeAssignment/takeCovalentActivity.do?locator=assignment-take hdlo weakest acid X G 2503(4) 2502o)- X O ChemlecSempt Content /adamson.bl.. A My presentations -. E Discord A Search results - Go. E Free Math & Scienc. O PracticePreCalculus. Paraphrasing Tool J. * OWLU2| Assignmen. Recordings for MM. P 07) Pinterest » E Reading I [References) One half-cell in a voltaic cell is constructed from a copper wire electrode in a 3.1 x 10 M solution of Cu(NO,)a The other half-cell consists of a zinc electrode n a 0.55 M selution of Zn(NO,),. Calculate the cell potential You may need to use the following data Zn+ (aq) + 2 e+ Zn(s) E-0.763 V Cu (aq) + 2 e Cu(s) E=+0.337 V Potential - UAV Submit Answer Retry Entire Group eq Not mast b=

Calculate AH for the reaction: 3C + 4H, -> CH; ΔΗ = ?? rxn Using the following data: AH = - 2220.1 KJ/mol 3C0, + 4H,0 02 -> Co, 50, --> AH = - 393.5 KJ/mol H2 1/202-> H,0 AH =- 285.8 KJ/mol a. -58.95 KJ/mol O b. +103.6 KJ/mol O c. -10.36 KJ/mol O d. -103.7 KJ/mol

Give a clear explanation answer....

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12. In the aqueous reaction below, the acid reactant is and its conjugate base product is CH;NH2 + HSO, 5 CH;NH;* + SO,²- CH3NH2; CH3NH3* b. CH;NH2; S0,2- c. HSO4"; CH3NH3* d. HSO,; SO,?- a. e. HSO4; H3O*