The balanced chemical equation and the expression for K a has to be written for the reaction between H 3 PO 4 and water. Concept introduction: Bronsted-Lowry acid will donate a proton to the water. The oxygen in the water molecule will have lone pair of electrons, which will accept the proton from the acid-forming hydronium ion H O 3 + . The acid dissociation constant gives the ratio of the product concentration and reactant concentration and the strength of the acid; that is, to what extent the acid dissociates in the solution.
The balanced chemical equation and the expression for K a has to be written for the reaction between H 3 PO 4 and water. Concept introduction: Bronsted-Lowry acid will donate a proton to the water. The oxygen in the water molecule will have lone pair of electrons, which will accept the proton from the acid-forming hydronium ion H O 3 + . The acid dissociation constant gives the ratio of the product concentration and reactant concentration and the strength of the acid; that is, to what extent the acid dissociates in the solution.
Solution Summary: The author explains that the balanced chemical equation and the expression for K_a has to be written for the reaction between Bronsted-Lowry acid and water.
Interpretation: The balanced chemical equation and the expression for Ka has to be written for the reaction between H3PO4 and water.
Concept introduction: Bronsted-Lowry acid will donate a proton to the water. The oxygen in the water molecule will have lone pair of electrons, which will accept the proton from the acid-forming hydronium ion HO3+ . The acid dissociation constant gives the ratio of the product concentration and reactant concentration and the strength of the acid; that is, to what extent the acid dissociates in the solution.
b)
Interpretation Introduction
Interpretation: The balanced equation and the expression for Ka has to be written for the reaction between C6H5COOH and water.
Concept introduction: Bronsted-Lowry acid will donate a proton to the water. The oxygen in the water molecule will have lone pair of electrons, which will accept the proton from the acid-forming hydronium ion HO3+ . The acid dissociation constant gives the ratio of the product concentration and reactant concentration and the strength of the acid; that is, to what extent the acid dissociates in the solution.
c)
Interpretation Introduction
Interpretation: The balanced equation and the expression for Ka must be written for the reaction between HSO4− and water.
Concept introduction: Bronsted-Lowry acid will donate a proton to the water. The oxygen in the water molecule will have lone pair of electrons which will accept the proton from the acid-forming hydronium ion HO3+ . The acid dissociation constant gives the ratio of the product concentration and reactant concentration and the strength of the acid; that is, to what extent the acid dissociates in the solution.
d)
Interpretation Introduction
Interpretation: The balanced equation and the expression for Ka has to be written for the reaction between HNO2 and water.
Concept introduction: Bronsted-Lowry acid will donate a proton to the water. The oxygen in the water molecule will have lone pair of electrons, which will accept the proton from the acid-forming hydronium ion HO3+ . The acid dissociation constant gives the ratio of the product concentration and reactant concentration and the strength of the acid; that is, to what extent the acid dissociates in the solution.
I have a excitation/emission spectra of a quinine standard solution here, and I'm having trouble interpreting it. the red line is emission the blue line is excitation. i'm having trouble interpreting properly. just want to know if there is any evidence of raman or rayleigh peaks in the spectra.
Give the major product of the following reaction.
excess
1. OH, H₂O
1.OH
H
CH3CH2CH21
H
2. A.-H₂O
Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and
Advanced Template toolbars. The single bond is active by default.
2. Use Hess's law to calculate the AH
(in kJ) for:
rxn
CIF(g) + F2(g) →
CIF 3 (1)
using the following information:
2CIF(g) + O2(g) →
Cl₂O(g) + OF 2(g)
AH = 167.5 kJ
ΔΗ
2F2 (g) + O2(g) → 2 OF 2(g)
2C1F3 (1) + 202(g) → Cl₂O(g) + 3 OF 2(g)
о
=
= -43.5 kJ
AH = 394.1kJ
Chapter 18 Solutions
Chemistry The Molecular Nature Of Matter And Change 9th
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