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2 years ago

Cuántos gramos de hidróxido de sodio (NaOH) se debe de agregar a un litro de agua para obtener una disolución al 25% en masa

Chemistry

1 answer:

Tamiku [17]2 years ago

5 0

Answer:

= 250 gramos

Explanation:

No ha preguntado sobre peso / peso o peso / volumen.

Para w / v:

Simplemente tome el 25% de 1000 y luego agréguelo en 1L de agua

La masa de NaOH necesaria es de 250 g.

Para w / w:

250 gramos de NaOH y 750 g de agua, hará que la solución sea 1L.

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A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o

Andreas93 [3]

<h3>Answer:</h3>

2.809 L of H₂SO₄

<h3>Explanation:</h3>

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

<h3>First: We need to write the balanced equation for the reaction.</h3>

The reaction between NaOH and H₂SO₄ is a neutralization reaction.
The balanced equation for the reaction is;

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

<h3>Second: We calculate the umber of moles of NaOH used </h3>

Number of moles = Mass ÷ Molar mass
Molar mass of NaOH is 40.0 g/mol

Therefore;

Moles of NaOH = 13.20 g ÷ 40.0 g/mol

= 0.33 moles

<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>

From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
Thus, Moles of H₂SO₄ = moles of NaOH × 2

= 0.33 moles × 2

= 0.66 moles of H₂SO₄

<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>

When given the molarity of an acid and the number of moles we can calculate the volume of the acid.

That is; Volume = Number of moles ÷ Molarity

In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

= 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

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Calculate the heat energy released when 21.1 g of liquid mercury at 25.00 °C is converted to solid mercury at its melting point.

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2 years ago

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2 years ago

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Calcium cyclamate, Ca(C₆H₁₁NHSO₃)₂, is an artificial sweetener used in many countries around the world but is banned in the Unit

xeze [42]

Answer:

2HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃;

Ba(C₆H₁₁NHSO₃)₂ + H₂SO₄ → BaSO₄ + 2HC₆H₁₁NHSO₃;

2HC₆H₁₁NHSO₃ + Ca(OH)₂ → Ca(C₆H₁₁NHSO₃)₂ + 2H₂O

Explanation:

First, let's see the reactants for the first reaction and how they dissociate:

HC₆H₁₁NHSO₃ → H⁺ + C₆H₁₁NHSO₃⁻

BaCO₃ → Ba²⁺ + CO₃²⁻ (Barium is from group 2, so its cation has charge +2)

So, to form the products, the cation of one will join the anion of others. The amount of the cation will be the charge of the anion, and the amount of the anion will be the charge of the cation:

H⁺ + CO₃²⁻ → H₂CO₃

Ba²⁺ + C₆H₁₁NHSO₃⁻ → Ba(C₆H₁₁NHSO₃)₂

The reaction then is:

HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃

The number of elements must be the same on both sides, so the balanced equation is

2HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃

The treatment with H₂SO₄ will produce:

H₂SO₄ → 2H⁺ + SO₄⁻²

Ba(C₆H₁₁NHSO₃)₂ → Ba²⁺ + C₆H₁₁NHSO₃⁻

The balanced reaction will be then:

Ba(C₆H₁₁NHSO₃)₂ + H₂SO₄ → BaSO₄ + 2HC₆H₁₁NHSO₃

In the last step, HC₆H₁₁NHSO₃ will react with Ca(OH)₂

HC₆H₁₁NHSO₃ → H⁺ + C₆H₁₁NHSO₃⁻

Ca(OH)₂ → Ca²⁺ + 2OH⁻

The balance reaction will be:

2HC₆H₁₁NHSO₃ + Ca(OH)₂ → Ca(C₆H₁₁NHSO₃)₂ + 2H₂O

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Oxygen and hydrogen form the polyatomic hydroxide ion. What is its charge?

prohojiy [21]

Answer:

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