None O Macmillan LearningFor the diprotic weak acid H₂A, Kal = 2.1 x 10-6 and Ka2 = 7.2 × 10-9.What is the pH of a 0.0600 M solution of H₂A?pH=3.95Incorrect AnswerWhat are the equilibrium concentrations of H2A and A² in this solution?7.2 x10-9[H₂A] =Incorrect Answer[A²-]=1.12 x10 4Incorrect AnswerMM

Answered: O Macmillan Learning For the diprotic…

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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[References] Calculate AH for the reaction: 2NH3 (g) + O2(g) → N2H4 (1) + H2O(1) given the following data: 2NH3 (g) + 3N2O(g) → 4N2(g) + 3H2O(1) AH =–1010. kJ N2O(9) + ЗH2(9) — N;Ha (1) + Hа0(1) ДН %3D — 317 kJ N2H4 (1) + O2 (9) → N2(g)+2H20(1) ΔΗ--623 k.J На (9) + О2(9) — Н,0(1) AH = -286 kJ ΔΗ- kJ Submit Answer Try Another Version 3 item attempts remaining
To balance the chemical equation Fe,O3 + CO → Fe + CO2, you could introduce coefficients a, b, c, d and write aFe20з + bCО — сFe + dCО2 then write linear equations for each element. The equation for Fe would be: The equation for O would be:
12
Use and modify some/all the reactions provided to form the indicated overall reaction according to Hess' Law. • If a reaction is not used in your Hess' Law process, include that indication by multiplying the reaction by "0" • If no modification is needed and the reaction is applied as written below, include that indication by multiplying the reaction by "1" • Use whole numbers and simple fractions (such as "2" or "1/2", not 0.5) and negative signs (such as "-2") where needed Then calculate and report the enthalpy change (in kJ) for the overall reaction in the indicated blank, including units with your written value and rounded to 1 decimal place. Overall Reaction: CS2(1) + 302(g) → CO2(g) + 2SO2(g) Rxn # Reaction ΔΗ° (kJ) 1 C(s) → C(g) 716.67 2 C(s) + O2(g) → CO2(g) -393.5 3 3Fe(s) +202(g) → Fe3O4(s) -1118 4 C(s) + 2S(s) → CS2(1) 87.9 5 CH4(g) + O2(g) → CO₂(g) + H₂O(g) -882.0 6 S(s) + O2(g) → SO2(g) -296.8 N2(g) + 4H2(g) + Cl2(g) → 2NH4Cl(g) -629.1 7
With the information provided answer C
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Balance the following equation. Rewrite the chemical equation and include the correct coefficients and the states of matter. Fe2O3(s) + CO(g) → Fe3O4(s) + CO2(g)
complete the following reaction
Is there a reason those sig figs are wrong or are my calculations just completely off?
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
(4d-202) A buffer solution is made by mixing H3PO4 and NaH2PO4 in water. How does this buffer system neutralize an addition of 5 mL of 0.45 M NaOH? Choose one of the options below. • The H3PO4, as a weak acid, will neutralize the added NaOH • The NaH2PO4, as a weak base, will neutralize the added NaOH • The H3PO4, as a weak base, will neutralize the added NaOH • The NaH2PO4, as a weak acid, will neutralize the added NaOH For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). Paragraph Arial 14px A v Ix Q x² X, g ST 田用国 EXE ABC Ť {;} +) P. O WORDS POWERED BY TINY +]
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.

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