Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card
Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card
15th Edition
ISBN: 9781119231318
Author: Morris Hein
Publisher: Wiley (WileyPLUS Products)
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Chapter 12, Problem 50PE

(a)

Interpretation Introduction

Interpretation:

Volume of H2 gas produced from 42.9 g Mg at STP has to be determined.

Concept Introduction:

Ideal gas law represented as equation to relate its volume and pressure with its temperature. Expression for ideal gas equation is as follows:

  PV=nRT        (1)

Here,

P is pressure of the gas.

V is volume of gas.

n denotes moles of gas.

R is gas constant.

T is temperature of gas.

(b)

Interpretation Introduction

Interpretation:

Number of moles of HCl required to produce 825 mL of H2 at STP has to be determined.

Concept Introduction:

Refer to part (a).

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Do not apply the calculations, based on the approximation of the stationary state, to make them perform correctly. Basta discard the 3 responses that you encounter that are obviously erroneous if you apply the formula to determine the speed of a reaction. For the decomposition reaction of N2O5(g): 2 N2O5(g) · 4 NO2(g) + O2(g), the following mechanism has been proposed: N2O5 -> NO2 + NO3_(K1) NO2 + NO3 →> N2O5 (k-1) → NO2 + NO3 → NO2 + O2 + NO (K2) NO + N2O5 → NO2 + NO2 + NO2 (K3) Give the expression for the acceptable rate. (A). d[N₂O] dt = -1 2k,k₂[N205] k₁+k₂ d[N₂O5] (B). dt =-k₁[N₂O₂] + k₁[NO2][NO3] - k₂[NO2]³ (C). d[N₂O] dt =-k₁[N₂O] + k₁[N205] - K3 [NO] [N205] (D). d[N2O5] =-k₁[NO] - K3[NO] [N₂05] dt
A 0.10 M solution of acetic acid (CH3COOH, Ka = 1.8 x 10^-5) is titrated with a 0.0250 M solution of magnesium hydroxide (Mg(OH)2). If 10.0 mL of the acid solution is titrated with 20.0 mL of the base solution, what is the pH of the resulting solution?
For the decomposition reaction of N2O5(g): 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed: N2O5 NO2 + NO3 (K1) | NO2 + NO3 → N2O5 (k-1) | NO2 + NO3 NO2 + O2 + NO (k2) | NO + N2O51 NO2 + NO2 + NO2 (K3) → Give the expression for the acceptable rate. → → (A). d[N205] dt == 2k,k₂[N₂O₂] k₁+k₁₂ (B). d[N2O5] =-k₁[N₂O] + k₁[NO₂] [NO3] - k₂[NO₂]³ dt (C). d[N2O5] =-k₁[N₂O] + k [NO] - k₂[NO] [NO] d[N2O5] (D). = dt = -k₁[N2O5] - k¸[NO][N₂05] dt Do not apply the calculations, based on the approximation of the stationary state, to make them perform correctly. Basta discard the 3 responses that you encounter that are obviously erroneous if you apply the formula to determine the speed of a reaction.

Chapter 12 Solutions

Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card

Ch. 12.8 - Prob. 12.11PCh. 12.9 - Prob. 12.12PCh. 12.9 - Prob. 12.13PCh. 12 - Prob. 1RQCh. 12 - Prob. 2RQCh. 12 - Prob. 3RQCh. 12 - Prob. 4RQCh. 12 - Prob. 5RQCh. 12 - Prob. 6RQCh. 12 - Prob. 7RQCh. 12 - Prob. 8RQCh. 12 - Prob. 9RQCh. 12 - Prob. 10RQCh. 12 - Prob. 11RQCh. 12 - Prob. 12RQCh. 12 - Prob. 13RQCh. 12 - Prob. 14RQCh. 12 - Prob. 15RQCh. 12 - Prob. 16RQCh. 12 - Prob. 17RQCh. 12 - Prob. 18RQCh. 12 - Prob. 19RQCh. 12 - Prob. 20RQCh. 12 - Prob. 21RQCh. 12 - Prob. 22RQCh. 12 - Prob. 23RQCh. 12 - Prob. 24RQCh. 12 - Prob. 25RQCh. 12 - Prob. 26RQCh. 12 - Prob. 1PECh. 12 - Prob. 2PECh. 12 - Prob. 3PECh. 12 - Prob. 4PECh. 12 - Prob. 5PECh. 12 - Prob. 6PECh. 12 - Prob. 7PECh. 12 - Prob. 8PECh. 12 - Prob. 9PECh. 12 - Prob. 10PECh. 12 - Prob. 11PECh. 12 - Prob. 12PECh. 12 - Prob. 13PECh. 12 - Prob. 14PECh. 12 - Prob. 15PECh. 12 - Prob. 16PECh. 12 - Prob. 17PECh. 12 - Prob. 18PECh. 12 - Prob. 19PECh. 12 - Prob. 20PECh. 12 - Prob. 21PECh. 12 - Prob. 22PECh. 12 - Prob. 23PECh. 12 - Prob. 24PECh. 12 - Prob. 25PECh. 12 - Prob. 26PECh. 12 - Prob. 27PECh. 12 - Prob. 28PECh. 12 - Prob. 29PECh. 12 - Prob. 30PECh. 12 - Prob. 31PECh. 12 - Prob. 32PECh. 12 - Prob. 33PECh. 12 - Prob. 34PECh. 12 - Prob. 35PECh. 12 - Prob. 36PECh. 12 - Prob. 37PECh. 12 - Prob. 38PECh. 12 - Prob. 39PECh. 12 - Prob. 40PECh. 12 - Prob. 41PECh. 12 - Prob. 42PECh. 12 - Prob. 43PECh. 12 - Prob. 44PECh. 12 - Prob. 45PECh. 12 - Prob. 46PECh. 12 - Prob. 47PECh. 12 - Prob. 48PECh. 12 - Prob. 49PECh. 12 - Prob. 50PECh. 12 - Prob. 51PECh. 12 - Prob. 52PECh. 12 - Prob. 53PECh. 12 - Prob. 54PECh. 12 - Prob. 55AECh. 12 - Prob. 56AECh. 12 - Prob. 57AECh. 12 - Prob. 58AECh. 12 - Prob. 59AECh. 12 - Prob. 60AECh. 12 - Prob. 61AECh. 12 - Prob. 62AECh. 12 - Prob. 63AECh. 12 - Prob. 64AECh. 12 - Prob. 65AECh. 12 - Prob. 66AECh. 12 - Prob. 67AECh. 12 - Prob. 68AECh. 12 - Prob. 69AECh. 12 - Prob. 70AECh. 12 - Prob. 71AECh. 12 - Prob. 72AECh. 12 - Prob. 73AECh. 12 - Prob. 74AECh. 12 - Prob. 75AECh. 12 - Prob. 76AECh. 12 - Prob. 77AECh. 12 - Prob. 78AECh. 12 - Prob. 79AECh. 12 - Prob. 80AECh. 12 - Prob. 81AECh. 12 - Prob. 82AECh. 12 - Prob. 83AECh. 12 - Prob. 84AECh. 12 - Prob. 85AECh. 12 - Prob. 86AECh. 12 - Prob. 87AECh. 12 - Prob. 88AECh. 12 - Prob. 89AECh. 12 - Prob. 90AECh. 12 - Prob. 91AECh. 12 - Prob. 92AECh. 12 - Prob. 93AECh. 12 - Prob. 94CECh. 12 - Prob. 95CECh. 12 - Prob. 96CECh. 12 - Prob. 97CECh. 12 - Prob. 98CECh. 12 - Prob. 99CECh. 12 - Prob. 100CECh. 12 - Prob. 101CE
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