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3 years ago

If 24.1 g of sodium hydroxide react with 22.0 g of hydrochloric acid to form 35.3g

Chemistry

1 answer:

disa [49]3 years ago

4 0

Answer:

10.85 g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

NaOH + HCl —> NaCl + H2O

Next, we shall determine the mass of NaOH that reacted and the mass of H2O produced from the balanced equation.

These can be obtained as illustrated below:

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

Mass of NaOH from the balanced equation = 1 × 40 = 40 g

Molar mass of H2O = (2×1) + 16 = 2 + 16 = 18 g/mol

Mass of H2O from the balanced equation = 1 × 18 = 18 g

Summary:

From the balanced equation above,

40 g of NaOH reacted to produce 18 g of H2O.

Finally, we shall determine the mass of water, H2O produced from the reaction as follow:

From the balanced equation above,

40 g of NaOH reacted to produce 18 g of H2O.

Therefore, 24.1 g of NaOH will react to produce = (24.1 × 18)/40 = 10.85 g of H2O.

Therefore, 10.85 g of H2O were produced from the reaction.

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The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.

maw [93]

CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L

7 0

3 years ago

Convert 3.30 g of copper (II) hydroxide Cu(OH)2 to molecules.

ratelena [41]

Answer:

0.18× 10²³ molecules

Explanation:

Given data:

Mass of copper hydroxide = 3.30 g

Number of molecules = ?

Solution:

Number of moles = mass/molar mass

Number of moles = 3.30 g/97.56 g/mol

Number of moles = 0.03 mol

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

0.03 mol × 6.022 × 10²³ molecules / 1mol

0.18× 10²³ molecules

6 0

2 years ago

When a solid changes to gas, what is it called??

Kay [80]

When solid changes it's phase to gas without forming liquid, process is called sublimation.

8 0

3 years ago

How many grams of carbon are present in 45.0g of CCl4

pentagon [3]

Molar Mass of CCl4= 12+4(35.5)=154 g/mol g Carbon= 45.0g/(154 g/mol)=0.292 mole

6 0

3 years ago

How many grams of NaF should be added to 500 mL of a 0.100 M solution of HF to make a buffer with a pH of 3.2

Korolek [52]

Answer:

2.25g of NaF are needed to prepare the buffer of pH = 3.2

Explanation:

The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:

pH = pKa + log [A-] / [HA]

<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>

The moles of HF are:

500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF

Replacing:

3.2 = 3.17 + log [A-] / [0.0500moles]

0.03 = log [A-] / [0.0500moles]

1.017152 = [A-] / [0.0500moles]

[A-] = 0.0500mol * 1.017152

[A-] = 0.0536 moles NaF

The mass could be obtained using the molar mass of NaF (41.99g/mol):

0.0536 moles NaF * (41.99g/mol) =

<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>

4 0

2 years ago