Construct the expression Kc for the following reaction. Sn(s) + 4 Cl:(g) = 2 SnCl-(1) 1 Drag the tiles into the numerator or denominator to form the expression. Each reaction participant must be represented by one tile. Do not combine terms. Kc 5 RESET [SNCL] [CL] [Sn] 2[SnCl] 4[Ch] 2[Sn] [SNCIJF [CLJ* [SnjF 2[SnClJ? 4[CL]* 2[Sn]?

Construct the expression Kc for the following reaction. Sn(s) + 4 Cl:(g) = 2 SnCl-(1) 1 Drag the tiles into the numerator or denominator to form the expression. Each reaction participant must be represented by one tile. Do not combine terms. Kc 5 RESET [SNCL] [CL] [Sn] 2[SnCl] 4[Ch] 2[Sn] [SNCIJF [CLJ* [SnjF 2[SnClJ? 4[CL]* 2[Sn]?

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...

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Consider this reaction: Br CH₂OH + Br-Br H₂CO The mechanism proceeds through a first cationic intermediate, intermediate 1. Nucleophilic attack leads to intermediate 2, which goes on to form final product. In cases that involve a negatively charged nucleophile, the attack of the nucleophile leads directly to the product. Year Hot Br + CH₂OH Br Intermediate 1 Intermediate 2 (product) In a similar fashion, draw intermediate 1 and intermediate 2 (final product) for the following reaction. Br H Br₂ + NaBr + H2O H H H NaOH . Pay attention to the reactants, they may differ from the examples. In some reactions, one part of the molecule acts as the nucleophile. Structures are provided for you to modify in the sketchers below. . • You do not have to consider stereochemistry. • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate intermediate 1 and intermediate 2 using the symbol from the dropdown menu. Sn [F H On [F I Br -OH HH…
⦁ 4. Make your own energy diagram for an exothermic reaction that does not include a catalyst and one with a catalyst (Be sure each line is a different color or labeled). Label the axis of the diagram. Then, state what a catalyst is and how a catalyst affects the reaction rate.
The following reaction energy diagram depicts another key atmospheric reaction. Label the axes, Ea (fwd), Ea (rev). and AH rxn, and transition state, and calculate the Ea (rev). * 20H 3 78 kJ 6 72 kJ O+ H20 2 1 = Potential energy; 2 = Reaction progress; 3 = Ea (fwd); 4 = transition state; 5 = Ea (rev); 6 =AH rxn; Ea (rev) = 6 kJ 1 = Potential energy; 2 = Reaction progress; 3 = Ea (rev); 4 = transition state; 5 = Ea (fwd); 6 =AH rxn; Ea (rev) = -6 kJ 1 = Reaction progress; 2 =Potential energy ; 3 = Ea (fwd); 4 = transition state; 5 = Ea (rev); 6 =AH rxn; Ea (rev) = 6 kJ 1 = Reaction progress; 2 =Potential energy ; 3 = Ea (rev); 4 = transition state; 5 = Ea (fwd); 6 =AH rxn; Ea (rev) = -6 kJ
47. Consider the reaction: MISSED THIS? Read Section 16.7; Watch KCV 16.7, IWE 16.7 NH4HS(s) NH3(g) + H₂S(8) At a certain temperature, Ke = 8.5 x 10-³. A reaction mixture at this temperature containing solid NH,HS has [NH3] = 0.166 M and [H₂S] = 0.166 M. Will more of the solid form or will some of the existing solid decompose as equilibrium is reached?
Construct the expression for Kc for the following reaction. Drag the tiles into the numerator or denominator to form the expression. Each reaction participant must be represented by one tile. Do not combine terms. PLOY JOHY 24.01 H.O (1) H (aq) + OH (aq) K₂ = HOP P+1 201 pry JOH RESET 201
When a chemical reaction is at equilibrium, the forward and reverse reactions occur at the same rate The concentrations of the different compounds in the reactions will remain constant. The relationship between the concentrations of the reactants and products is expressed as the equilibrium constant, Ke Drag each item to the appropriate bin. ▸View Available Hint(s) [0₂] Numerator Submit [H₂O]² [HC] [HCI] [Cl₂] Denominator 10₂1² [H₂O] Not used Reset Help [Cl₂1²
Use the References to access important values if needed for this question. Write the Ksp expression for the sparingly soluble compound calcium phosphate, Ca3 (PO4)2. (If either the numerator or denominator is 1, please enter 1.) Ksp Submit Answer Retry Entire Group 9 more group attempts remaining Previous Next
Label each reactant and product in the given chemical reaction. R F V CH4 96 5 G Search or type URL T + 20₂ G B tv 6 Y H Answer Bank product MacBook Pro N 9 U J CO₂ reactant 00 M 1 K ģ 69 + I O 2H₂0 L O command P Λ' : ; I option { [ ? = 1 O I } 1 dele
Construct the expression for Kc for the following reaction. 2 Sb(s) + 5 CL(g) =2 SbCls(g) Drag the tiles into the numerator or denominator to form the expression. O RESET 2[SDCL] 2[SbClj? [CL] 5[CL] [Sb] 2[Sb] [SbF 2[Sb] 46°F a earch 近
Construct the expression for Kc for the following reaction. 2 P(s) + 3 Cl(g) 2 PCI (g) Drag the tiles into the numerator or denominator to form the expression. Each reaction participant must be represented by one tile. Do not combine terms. [PCL) Ke (PP ICI. Incorrect, 2 attempts remaining Your expression for Kc is incorrect. Recall the general definition of Kc: aA + bB cC + dD Kc = [C]^c [D]^d/ [A]^a [B]^b
Based on your ICE table (Part 1) and the definition of Ka, set up the expression for Ka in order to determine the unknown. Each reaction participant must be represented by one tile. Do not combine terms. Ka = = 1.8 × 10-5 [500.0] [0.200] RESET [5.00] [1.0 × 10-9] [1.0 × 10-9] [1.8 × 10-5] [x + 5.00] [x + 1.0 × 10] [x-1.0 x 10 x + 1.0 × 10-] [x-1.0 × 10-] [x+1.8 × 10] [x-1.8 × 10%] [0] [x-5.00] < PREV 1 2 3 Based on your ICE table (Part 1) and the equilibrium expression, Ka (Part 2), determine the original mass of solid CH3COONa added to create this buffer solution. MNACH3COO g 0 0.360 30 21 43 11 15 36 x 10' 59 x 105 43 x 105 RESET 59 50
Consider the quilibrium reaction between X and Y, as shown below: X=Y AG The reaction is started with 10 mmol of X; no Y is initially present. After 48 hours, analysis reveals the presence of 10 mmol of X and 0 mmol of Y. Which is the most likely explanation? = −1 - 45 kJ mol X and Y have reached equilibrium concentrations. An enzyme has shifted the equilibrium toward X. Formation of Y is kinetically slow; equilibrium has not been reached by 48 hours. Formation of Y is thermodynamically unfavorable. Two of the above explanations are reasonable.