equilibrium. You can leave out water tself Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the bases row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that NH, is a weak base. O cids O LI mol of HI is added to 1.0 Lof a 1.1M NH, O bases: 0 solution Oother 0.09 moi of KOH is added to Oacids: 0 1OL of a solution that is 0.5M in both NH, and O bases O NH,CL. O other O please answer complete question or better to leave| D OO

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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equilibrium. You can leave out water tself.
Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the
formulas of the species that will act as neither acids nor bases in the 'other' row.
You will find it useful to keep in mind that NH, is a weak base.
O acids
O an
LI mol of HI is added to
1.0 L of a 1.1M NH,
O bases: 0
solution.
Oother 0
0.09 mol of KOH is added to acids: U
10L of a solution that is
0.5M in both NH, and
O bases O
NH,CI.
O other O
please answer complete question or better to leave
Transcribed Image Text:equilibrium. You can leave out water tself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that NH, is a weak base. O acids O an LI mol of HI is added to 1.0 L of a 1.1M NH, O bases: 0 solution. Oother 0 0.09 mol of KOH is added to acids: U 10L of a solution that is 0.5M in both NH, and O bases O NH,CI. O other O please answer complete question or better to leave
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