A 30.0 mL volume of 0.12 M NaOH is required to neutralize 20.0 ml of a vinegar solution.Calculate the morality of acetic acid in the vinegar solution?Select one:О a. 0.36 МОb. 0.18 МО с. 0.24 МO d. 3.6 M Write the Net ionic equation for the equation:Nacl + AgNO3 → AgCl + NANO3o a. Na*(aq) + Cl (aq) + Ag*(aq) + NO3 (aq) → AgCl(s) + Na*(aq) + NO3b.Na*(aq) + Cl (aq)+ Ag*(aq) + NO3 (aq) → Ag*(aq) + Cl(aq) + Na*(aq) + NO3O c.Cl(aq) + Ag*(aq) → AgCl(s)Od.Na*(aq) +NO3 (aq) → NaNO3 (s)

Since you have posted multiple questions, the answer for first question is given below. kindly…

A 30.0 mL volume of 0.12 M NAOH is required to neutralize 20.0 ml of a vinegar solution. Calculate the morality of acetic acid in the vinegar solution? Select one: О а. 0.36 M Оъ. 0.18 М О с. 0.24 М O d. 3.6 M

A 30.0 mL volume of 0.12 M NAOH is required to neutralize 20.0 ml of a vinegar solution. Calculate the morality of acetic acid in the vinegar solution? Select one: О а. 0.36 M Оъ. 0.18 М О с. 0.24 М O d. 3.6 M

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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When a chemical species is transformed into another chemical species it is said to have undergone a chemical reaction. It consists of breaking existing bonds and forming new bonds by changing the position of electrons. These reactions are best explained using a chemical equation.

Question

A 30.0 mL volume of 0.12 M NaOH is required to neutralize 20.0 ml of a vinegar solution.
Calculate the morality of acetic acid in the vinegar solution?
Select one:
О a. 0.36 М
Оb. 0.18 М
О с. 0.24 М
O d. 3.6 M

Write the Net ionic equation for the equation:
Nacl + AgNO3 → AgCl + NANO3
o a. Na*(aq) + Cl (aq) + Ag*(aq) + NO3 (aq) → AgCl(s) + Na*(aq) + NO3
b.
Na*(aq) + Cl (aq)
+ Ag*(aq) + NO3 (aq) → Ag*(aq) + Cl(aq) + Na*(aq) + NO3
O c.
Cl(aq) + Ag*(aq) → AgCl(s)
Od.
Na*(aq) +NO3 (aq) → NaNO3 (s)

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