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3 years ago

Esterification is what type of reaction? 1.addition 2.neutralization 3.combustion 4.equilibrium

Chemistry

2 answers:

Oksi-84 [34.3K]3 years ago

6 0

Answer:

Equilibrium reaction

Explanation:

shepuryov [24]3 years ago

5 0

Answer:

Esters and water are formed when alcohols react with carboxylic acids.

Explanation:

This reaction is called esterification, which is a reversible reaction. This type of reaction is called a condensation reaction, which means that water molecules are eliminated during the reaction. hope this help you :)

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What is the empirical formula of a compound that contains 25.92 percent nitrogen and 74.07 percent oxygen?

pychu [463]

N2O5

Explanation!

When given %, assume you have 100 g of the substance. Find moles, divide by lowest count. In this case you'll end up with

25.92 g N14.01 g N/mol N=1.850 mol N

74.07 g O16.00 g O/mol O=4.629 mol O

The ratio between these is 2.502 mol O/mol N, which corresponds closely with N2O5.

4 0

3 years ago

Read 2 more answers

Aluminum chloride can be formed from its elements:

saul85 [17]

Answer: The $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of $AlCl_3$ is:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)$ $\Delta H^o_{formation}=?$

The intermediate balanced chemical reaction are:

(1) $HCl(g)\rightarrow HCl(aq.)$ $\Delta H_1=-74.8kJ$ ( × 6)

(2) $H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$ $\Delta H_2=-185kJ$ ( × 3)

(3) $AlCl_3(aq.)\rightarrow AlCl_3(s)$ $\Delta H_3=+323kJ$ ( × 2)

(4) $2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)$ $\Delta H_4=-1049kJ$

The expression for enthalpy of formation of $AlCl_3$ is,

$\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]$

Putting values in above equation, we get:

$\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ$

Hence, the $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

6 0

3 years ago

For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a

Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁ -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁ ----> P cancels out

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

7 0

3 years ago

The flotation process used in metallurgy involves Multiple Choice the roasting of sulfides. separation of gangue from ore. elect

densk [106]

The flotation process used in metallurgy involves the separation of gangue from ore.

<h3>How does the flotation process work?</h3>

The various wettability qualities of a material's surface are the foundation of flotation operations. The basic principles of flotation are quite similar to those of a sink and float process, where the materials' relative densities to the medium in which they are deposited determine the basis of the separation.

<h3>What is the process of separating minerals from gangue known as?</h3>

Mineral processing, mineral dressing, or ore dressing are all terms for the process of separating minerals from gangue. It is an important and frequently necessary part of mining. Depending on the type of minerals used, the process may be difficult.

Learn more about minerals here:-

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8 0

2 years ago

What is the difference between the small and large intestines?

nalin [4]

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3 0

3 years ago