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3 years ago

Determine whether each phrase describes carboxylic acids or esters.

Chemistry

1 answer:

Thepotemich [5.8K]3 years ago

8 0

Answer:

Carboxylic acids produce hydrogen bonds amongst themselves and possess lower vapor pressure. They generally possess a sour odor. When an acid and a base react with each other to produce salt and water and comprises the combination of hydrogen and hydroxide ions, the reaction is termed the neutralization reaction. Thus, when carboxylic acid reacts with base the reaction is termed neutralization.

On the other hand, esters are known for their pleasant fragrances. They do not produce hydrogen bonds amongst themselves and possess higher vapor pressure. A hydration reaction in which free hydroxide dissociates the ester bonds between the glycerol and fatty acids of a triglyceride, leading to the formation of free fatty acids and glycerol is termed saponification.

Thus, the given blanks can be filled with carboxylic acid, carboxylic acid, esters, esters, esters, and carboxylic acid.

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In the following net ionic equation, identify each reactant as either a Bronsted-Lowry acid or a Bronsted-Lowry base. HCN(aq) H2

vfiekz [6]

Answer:

Explanation:

The definition of acids and bases by Arrhenius Theory was modified and extended by Bronsted-Lowry.

Bronsted-Lowry defined acid as a molecule or ion which donates a proton while a base is a molecule or ions that accepts the proton. This definition can be extended to include acid -base titrations in non-aqueous solutions.

In this theory, the reaction of an acid with a base constitutes a transfer of a proton from the acid to the base.

From the given information:

$\mathsf{HCN _{(aq)} + H_2O_{(l)} \to CN^{-}_{(aq)} + H_3O_{(aq)}}$

From above:

We will see that HCN releases an H⁺ ion, thus it is a Bronsted-Lowry acid

$H_2O$ accepts the H⁺ ion ,thus it is a Bronsted-Lowry base.

The formula of the reactant that acts as a proton donor is <u>HCN</u>

The formula of the reactant that acts as a proton acceptor is <u>H2O</u>

8 0

3 years ago

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are

VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell = 0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0

3 years ago

Is the process represented below an example of a physical or a chemical

Basile [38]

It is a physical change because only the states as being changes, not the actual bonds in the compound.

6 0

3 years ago

Consider the following unbalanced particulate representation of a chemical equation:

nevsk [136]

Answer:

H2 + I2 --> 2HI

Explanation:

The two reactants are diatomic molecules because they contain two atoms of the same element. Therefore, they would need to have a subscript of "2" next to their symbols.

When balancing an equation, you want the same amount and type of atoms on both sides. By adding a coefficient of "2" in front of the product, two H's and two I's are now on both sides.

3 0

3 years ago

Can someone help. posted this same question 3-4 times.

olya-2409 [2.1K]

(JSYK: this is all considering that only whole batches are included and fractional batches aren't)

Answer: 3 batches (72 cookies), her limiting reactant should be sugar since she only has enough to make three batches exactly, and she will also have 3/4 a cup of flour left over, 2 cups butter, and 5 eggs, in addition to the excess of baking soda and baking powder.

Hope this helps, if not, comment below please!!!

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3 years ago