None 1. Determine the pH (to two decimal places) of thefollowing solutions. A table of pKa values can be foundhere. If the 5% approximation is valid, use theassumption to compute pH.5.05×10-2 M propanoic acid 3.08Computer's answer now shown above.You are correct.Your receipt no. is 167-7882 ?PreviousTries9.86×10-1 M methylammonium chloride5.82Submit Answer Incorrect. Tries 1/5 Previous Tries7.77×10-3 M sodium acetate 8.82Submit AnswerIncorrect. Tries 1/5 Previous Tries8.13×103 M sodium acetylsalicylate9.80Submit Answer Incorrect. Tries 1/5 Previous Tries

Please take a look at the attached photo for the solution for each problem.

Answered: 1. Determine the pH (to two decimal…

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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Rb₂O(s) +____H₂O(1)→ Co2(CO3)3(s) + heat → 3_H₂O(1) + 1_As2O5(s) →➜ NaClO3(s) + heat → (110q000 sd, opt! 2 H3 As 0, 4 Ini Siyoda canl Mg(s) + H₂SO4(aq) → Myson noileupo unor ton boonstad to out + H₂ Om adami zogasdo ne
t- U- S- 요 s O NHẠCH3, H® NHE+₂, HⓇ CN Ⓒ ΝΗ, Η
ORT SHEET Heat of Neutralization EXPERIMENT elemsboomfchg lom 12 A. Heat Capacity of Calorimeter 1. Temp. of calorimeter and water before mixing 2. Temp. of warm water °C 22.0 39,0 30.3 3. Maximum temp. determined from your curve °C 4. Heat lost by warm water (temp decrease x °C 50.0 g x 4.184 J/K-g) = 02), 5. Heat gained by cooler water (temp. increase x 50.0 g x 4.184 J/K-g) = 30,3 22.0)x 13626J s0.0gmpi S0.0gy 6. Heat gained by the calorimeter [(4) – (5)] = 7. Heat capacity of calorimeter: heat gained by the calorimeter temperature increase J/K 3. Heat of Neutralization of HCl-NaOH 22.2 22.2. °C . Temp. of calorimeter and NaOH Temp. of HCI AT determined from your curve after adding HC1 °C to the NaOH Heat gained by solution (temperature increase x ON 100 g x 4.184 J/K-g) = 9977.8J %3D Heat gained by calorimeter (temperature increase x heat capacity of calorimeter) = J %3D Total joules released by reaction [(3) + (4)] = Tight O 2018 Pearson Education, Inc.
2. Write the expression for Kc and Kp if applicable, for the following reversible reactions at equilibrium: (a) HF(aq) + H;0() H,0 (aq) + F (aq) (b) 2NO(g) + 0,(g) 2NO0,(g) (C) CH,COOH(aq) + C,H,OH(aq) CH,C0OC,H,(aq) + H,O()
What are the basic requirements for a calirometer?
[References] Given the following data СэНа (9) + 302(9) — 2 СО2(9) + 2Н,0() 2C2 H6 (g) + 702(9) → 4 CO2(g) + 6H2O(1) 2H2 (g) + O2 (g) → 2H2 O(1) ΔΗ -1411.0 kJ AH = -3119.8 kJ AH=-571.7 kJ calculate A H for the reaction C2H4 (9) + H2 (g) → C2H6 (9) ΔΗΞ kJ Submit Answer Try Another Version 3 item attempts remaining
CI Cl2 heat 8⁰ (1) (2) (3) Based on your results for the radical chlorination experiment, predict which products would be most and least abundant in the product mixture. Briefly justify your answer. CI
Ag(s) + PbSO(s) → Pb(s) + Ag* (ac) + So²² (ac). Datos AH 298: 53, 22 Kcal :53,41cal 4 AG 298: Cal дор, 8, 7- 298: K InKT K(T) = ? = InK 1 55,8 8,7 x 10 + 298 R -dᎢ . 298 T TE F 2,81 x 10 =28,8- 4,4lnT. T
▼ Part B 2NO3(aq) +8H(aq) + 6e 3Fe(s) + 2NO3 (aq) +8H+ (aq) 3Fe(s)→→3Fe²+ (aq) + 6e →>> 2NO(g) + 4H₂0 (1) →3Fe²+ (aq) + 2NO(g) + 4H₂O(1) IVE ΑΣΦ Calculate El using the tabulated standard electrode potentials at 25 °C. cell Express your answer in volts to three significant figures. Eeell= 1.21 SWE Submit Previous Answers Request Answer 2+ X Incorrect; Try Again; 4 attempts remaining ? V stontiol buoubtrooting the electro na
500.0 mL of 0.100 mol·L HCl(aq) was added to a high quality insulated calorimeter containing 500.0 mL of 0.100 mol·LNaOH(aq). Both solutions had a density of 1.000 g·mL and a specific heat of 4.184 J·g ·K . The calorimeter had a heatcapacity of 850.0 J·K . The temperature of the entire system rose from 25.50 C to 26.05 C.Calculate the heat of reaction. Thank you!
Name these organic compounds: Η Η Η Η Η Η ||| C=C=C-H ||| Η Η Η Η Η II H — C =C=C=C=C - H ||||| Η structure Η Η Η Η Η Η H H Η Η Η | Η ||| H – C – C =C=C=C=C=CH ||||||| Η Η Η Η Η Η Η name Π Π Π

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