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siniylev [52]
3 years ago
10

A 25.00 ml solution of sulfuric acid H2SO4 is titrated to phenolphthalein end point with 27.00 ml of 1.700 M KOH

Chemistry
1 answer:
Nastasia [14]3 years ago
5 0
<h3>Answer:</h3>

0.918 M

<h3>Explanation:</h3>

Assuming the question requires we calculate the Molarity of sulfuric acid:

We are given:

  • Volume of the acid, H₂SO₄ = 25.00 ml
  • Volume of the base, KOH = 27.00 mL
  • Molarity of the base, KOH is 1.70 M

We can calculate the molarity of the acid using the following steps;

<h3>Step 1: Write the chemical equation for the reaction.</h3>

The reaction is an example of a neutralization reaction where a base reacts with an acid to form salt and water.

Therefore, the balanced equation will be;

H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)

<h3>Step 2: Determine the moles of the base, KOH </h3>

When given molarity and the volume of a solution, the number of moles can be calculated by multiplying molarity with volume.

Number of moles = Molarity × Volume

                             = 1.700 M × 0.027 L

                              = 0.0459 moles

Thus, moles of KOH used is 0.0459 moles

<h3>Step 3: Determine the number of moles of the Acid, H₂SO₄</h3>

From the reaction, 1 mole of the acid reacts with 2 moles of KOH

Therefore, the mole ratio of H₂SO₄ to KOH is 1 : 2

Thus, moles of H₂SO₄ = Moles of KOH ÷ 2

                                     = 0.0459 moles ÷ 2

                                     = 0.02295 moles

<h3>Step 4: Calculate the molarity of the Acid </h3>

Molarity is the concentration of a solution in moles per liter

Molarity = Moles ÷ Volume

Molarity of the acid = 0.02295 moles ÷ 0.025 L

                                = 0.918 M

Thus, the molarity of the acid, H₂SO₄ is 0.918 M

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In-s [12.5K]

The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of  3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm

<h3>What is Mole ?</h3>

A mole is a very important unit of measurement that chemists use.

A mole of something means you have 6.023 x 10 ²³ of that thing.

  • For 2.15 mol of hydrogen sulphide (H₂S) :

1 mole hydrogen sulphide (H₂S) = 34.08088 grams

Therefore,

2.15 mol of hydrogen sulphide (H₂S) = 34.08088 grams x 2.15 mol

                                                              = 73.272 gm

  • For 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) ;

1 mol of lead(II) iodide, (PbI₂) = 461.00894 grams

Therefore,

3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) = 461.00894 grams x 3.95 × 10⁻³ mol

                                                                  = 1.82 gm

Hence,The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of  3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm

Learn more about mole here ;

brainly.com/question/21323029

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When performing labratory experiments, which of the following items is ALWAYS necessary to use?
densk [106]
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Do you think that the human being is the center of the universe?
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3 0
2 years ago
How many molecules in each sample?<br><br> 64.7 g N2<br> 83 g CCl4<br> 19 g C6H12O6
lilavasa [31]

Answer:

  • 1.39x10²⁴ molecules N₂
  • .25x10²³ molecules CCl₄
  • 6.38x10²² molecules C₆H₁₂O₆

Explanation:

First we <u>convert the given masses into moles</u>, using the <em>compounds' respective molar mass</em>:

  • 64.7 g N₂ ÷ 28 g/mol = 2.31 mol N₂
  • 83 g CCl₄ ÷ 153.82 g/mol = 0.540 mol CCl₄
  • 19 g C₆H₁₂O₆ ÷ 180 g/mol = 0.106 mol C₆H₁₂O₆

Then we multiply each amount by <em>Avogadro's number</em>, to <u>calculate the number of molecules</u>:

  • 2.31 mol N₂ * 6.023x10²³ molecules/mol = 1.39x10²⁴ molecules
  • 0.540 mol CCl₄ * 6.023x10²³ molecules/mol = 3.25x10²³ molecules
  • 0.106 mol C₆H₁₂O₆ * 6.023x10²³ molecules/mol = 6.38x10²² molecules
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2 years ago
State the definition of the partial molar Gibbs energy.
balu736 [363]

Explanation :

As we know that the Gibbs free energy is not only function of temperature and pressure but also amount of each substance in the system.

G=G(T,P,n_1,n_2)

where,

n_1\text{ and }n_2 is the amount of component 1 and 2 in the system.

Partial molar Gibbs free energy : The partial derivative of Gibbs free energy with respect to amount of component (i) of a mixture when other variable (T,P,n_j) are kept constant are known as partial molar Gibbs free energy of i^{th} component.

For a substance in a mixture, the chemical potential (\mu) is defined as the partial molar Gibbs free energy.

The expression will be:

\bar{G_i}=\mu_i=\frac{\partial G}{\partial n_i}_{(T,P,n_j)}

where,

T = temperature

P = pressure

n_i\text{ and }n_j is the amount of component 'i' and 'j' in the system.

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2 years ago
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