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Ket [755]
3 years ago
15

You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH sol

ution to exactly neutralize this sample (phenolphthalein was used as an indicator). What is the molarity of the acetic acid solution? What is the percentage of acetic acid in the solution?
Chemistry
2 answers:
oee [108]3 years ago
6 0

Answer:

0.2788 M

1.674 %(m/V)

Explanation:

Step 1: Write the balanced equation

NaOH + CH₃COOH → CH₃COONa + H₂O

Step 2: Calculate the reacting moles of NaOH

0.03575 L \times \frac{0.1950mol}{L} = 6.971 \times 10^{-3} mol

Step 3: Calculate the reacting moles of CH₃COOH

The molar ratio of NaOH to CH₃COOH is 1:1.

6.971 \times 10^{-3} molNaOH \times \frac{1molCH_3COOH}{1molNaOH} = 6.971 \times 10^{-3} molCH_3COOH

Step 4: Calculate the molarity of the acetic acid solution

M = \frac{6.971 \times 10^{-3} mol}{0.02500L} =0.2788 M

Step 5: Calculate the mass of acetic acid

The molar mass of acetic acid is 60.05 g/mol.

6.971 \times 10^{-3} mol \times \frac{60.05g}{mol} =0.4186 g

Step 6: Calculate the percentage of acetic acid in the solution

\frac{0.4186g}{25.00mL}  \times 100\% = 1.674 \%(m/V)

Ksju [112]3 years ago
4 0

Answer:

Concentration  acetic acid = ‬0.27885 M

% acetic acid = 0.69%

Explanation:

You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH solution to exactly neutralize this sample (phenolphthalein was used as an indicator). What is the molarity of the acetic acid solution?

what is the percentage of acetic acid in the solution? Assume the density of the solution is 1 g/ml.

Step 1: Data given

Volume of acetic acid = 25.00 mL = 0.025 L

Volume of NaOH = 35.75 mL = 0.03575 L

Molarity of NaOH = 0.1950 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

Moles = molarity * volume

Moles NaOH = 0.1950 M * 0.03575 L

Moles NaOH = 0.00697125‬ moles

Step 4: Calculate concentration of acetic acid

We need 0.00697125‬ moles of acetic acid to neutralize NaOH

Concentration = moles / volume

Concentration = 0.00697125 moles / 0.025 L

Concentration = ‬0.27885 M

Step 5: Calculate mass of acetic acid

Mass acetic acid = moles * molar mass

Mass acetic acid = 0.00697125 moles * 60.05g/mol

Mass acetic acid = 0.4186 grams

Step 6: Calculate mass of sample

Total volume = 60.75 mL = 0.06075 L

Mass of sample 60.75 mL * 1g/mL = 60.75 grams

Step 7: Calculate the percentage of acetic acid in the solution

% acetic acid = (0.4186 grams / 60.75 grams ) * 100 %

% acetic acid = 0.69%

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Certain compound contains 7.3% carbon, 4.5% hydrogen, 36.4% oxygen, and 31.8% nitrogen. It’s reality molecular mass is 176.0. Fi
erastovalidia [21]

Answer:

The answer to your question is:

Explanation:

Data

carbon        7.3%          =     7.3g

hydrogen    4.5%         =      4.5g

oxygen       36.4%         =     36.4 g

nitrogen     31.8%         =     31.8 g

Now

For carbon

                    12 g --------------------1 mol

                    7.3 g     -------------     x

                       x = 7.3/12 = 0.608 mol

For hydrogen

                 1 g   --------------------  1 mol

                 4.5 g  ------------------    x

                   x = 4.5 mol

For oxygen

             16 g ------------------- 1 mol

             36.4 g ----------------    x

             x = 2.28 mol

For nitrogen

              14 g   ----------------   1 mol

              31.8 g ---------------    x

             x = 2.27 mol

Now divide by the lowest result, the is 0.608 from carbon

carbon              0.608/0.608 = 1

hydrogen           4.5/ 0.608 = 7.4

oxygen              2.28/0.608 = 3.75

nitrogen             2.27/0.608 = 3.73

Empirical formula = CH₇O₄N₄

     

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2 years ago
When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for
balu736 [363]

Answer:

1.

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-

Whereas the half reactions are:

Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O

Next, we exchange the transferred electrons:

1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow  4N^{4+}O^{2-}_2+4H_2O

Afterwards, we add them to obtain:

Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O

By adding and subtracting common terms we obtain:

Sn^0+4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O

Finally, by removing the oxidation states we have:

Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

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Is a solid because liquids are not is not that close and gas is very far apart.
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Answer:

0.159 M

Explanation:

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\frac{M1V1}{V2} = M2

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