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2 years ago

Why does benzocaine precipitate during neutralization?.

Chemistry

1 answer:

Natalija [7]2 years ago

5 0

Explain why benzocaine precipitates during the neutralization.

With the addition of an acid, the amine group on the benzocaine will become protonated. The protonation creates a positive charge on the benzocaine molecule. Because of this, benzocaine will become soluble in a polar solvent. However, when the solution is returned to a more basic pH level, the amine will no longer be protonated. Thus, the benzocaine will no longer be soluble and will precipitate during neutralization.

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What’s the equation to solid calcium oxide reacts with hydrochloric acid to form a solution of calcium chloride and water as it’

meriva

Answer: CaO + 2HCl —> CaCl2 + H2O

Explanation:

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3 years ago

PLEASE HELPPPP How would life on earth be different if it axis were not tilted which respect to its orbit

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3 years ago

Read 2 more answers

What is the slope of the line?

pychu [463]

Answer:

The slope of the line is $-\frac{10}{7}$ .

Explanation:

The slope of the line ( $m$ ) is the change in dependent variable ( $y$ ) divided by the change in independent variable ( $x$ ):

$m = \frac{y_{f}-y_{o}}{x_{f}-x_{o}}$ (1)

If we know that $(x_{o}, y_{o}) = (0, 2)$ and $(x_{f}, y_{f}) = (1.4, 0)$ , then the slope of the line is:

$m = \frac{0-2}{1.4-0}$

$m = -\frac{10}{7}$

The slope of the line is $-\frac{10}{7}$ .

3 0

2 years ago

Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0

3 years ago

What has the lowest ionization erergy?

Svetach [21]

Answer:

Francium (Fr)

Explanation:

From the given choices, francium will have the lowest ionization energy.

Ionization energy is the energy required to remove the most loosely held electron within an atom.

The magnitude of the ionization energy depends on the characteristics of the atom in relation to its nuclear charge, atomic radius, stability etc.

Generally on the periodic table, ionization energy increases from left to right on the table
As you go from metals to non-metals and to gases, the value of the ionization energy increases steadily.
Down the group, the value reduces.
Since Francium is the most metallic of all the given choices, it has the highest ionization energy.

3 0

3 years ago