All categories

3 years ago

What is the product of the reaction of pentanoic acid with ethanol in the presence of a strong acid?

Chemistry

1 answer:

balandron [24]3 years ago

7 0

Answer:

ethylpentanoate

Explanation:

Alkanoic acids react with alkanols in the presence of mineral acids to yield an ester and water. This is the organic analogue of the inorganic neutralization reaction. The reaction his commonly called esterification. It is an acid catalysed reaction.

The reaction of pentanoic acid and ethanol in the presence of a string acid is shown below;

CH3CH2CH2CH2COOH(aq) + CH3CH2OH(aq) ----> CH3CH2CH2CH2COOCH2CH3(aq) + H2O(l)

The name of the compound formed is ethylpentanoate.

You might be interested in

The melting point of a substance is a ________ property of the substance

4vir4ik [10]

Chemical property of the substance but then again it could be physical

8 0

3 years ago

One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 77°C. Weights are remo

frozen [14]

Answer:

The total work is 4957.45J

Explanation:

For an ideal gas, at constant temperature the definition of work (W) is

$W = - P.dV = - n.R.T \int\limits^i_f {\frac{dV}{V} \\W = - n.R.T. Ln (\frac{V_{f}}{V_{I}})\\W = - n.R.T. Ln (\frac{P_{i}}{P_{f}})$

where P is the pressure, V the volume, n the moles number, T the temperature and R the gas constant.

To solve the problem is necessary to replace the two steps in the equation

Stape 1: n = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 5.50atm and Pf = 2.43atm.

$W_{1} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{5.50atm}{2.43atm}) = 23.44atm.Lx101.325\frac{J}{atm.L} =2375.44J$

Stape 2: n = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 2.43atm and Pf = 1.00atm.

$W_{2} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{2.43atm}{1.00atm}) = 25.48atm.Lx101.325\frac{J}{atm.L} =2582.01J$

The total work is the sum of the two steps

$W = W_{1} + W_{2} = 2375.44J + 2582.01J = 4957.45J$

4 0

3 years ago

Which of the following is NOT true of protein quaternary structure?

AVprozaik [17]

The choices for this question can be found elsewhere and as follows:

A. A single polypeptide may have quaternary structure.
B. Hydrogen bonds may hold the polypeptides in contact.
C. A quaternary protein cannot have fewer than two carboxyl groups.

I think the correct answer from the choices is option A. The statement a single polypeptide may have quaternary structure is not true.

3 0

3 years ago

If 345.1 g of CO2 are placed in a vessel whose volume is 32.1 L at a temperature of 20.0oC, what will the pressure be? (R= 0.082

Finger [1]

Answer:

5.88atm

Explanation:

First, we obtain the number of mole of CO2 present in the vessel. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the question = 345.1g

Number of mole of CO2 =?

Number of mole = Mass/Molar Mass

Number of mole of CO2 = 345.1/44

= 7.84moles

Now we can easily calculate the pressure by doing the following:

Data obtained from the question include:

V (volume) = 32.1 L

T (temperature) = 20°C = 20 + 273 = 293K

R (gas constant) = 0.0821atm*L/mole*K

n (number of mole) = 7.84moles

P (pressure) =?

We will be making use of the ideal gas equation PV = nRT to calculate the pressure

PV = nRT

P = nRT/V

P = 7.84 x 0.0821 x 293/32.1

P = 5.88atm

Therefore, the pressure is 5.88atm

7 0

3 years ago

An aqueous solution containing 9.82 g9.82 g of lead(II) nitrate is added to an aqueous solution containing 5.76 g5.76 g of potas

n200080 [17]

Answer:

The limiting reactant is lead(II) nitrate.

7.20 g of precipitate are formed.

1.9 g of the excess reactant remain.

Explanation:

The reaction that takes place is:

Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

With a percent yield of 87.5%.

To determine the limiting reactant, first we convert the masses of each reactant to moles, using their molar mass:

9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂

5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl

Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.

To calculate the mass of precipitate (PbCl₂) formed, we use the moles of the limiting reactant:

0.0296 mol Pb(NO₃)₂ $\frac{1molPbCl_{2}}{1molPb(NO_{3})_{2}}$ * $\frac{278.1g}{1molPbCl_{2}}$ * 87.5/100 = 7.20 g PbCl₂

- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:

0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂

Now we convert that amount to moles of KCl and finally into grams of KCl:

0.0259 mol Pb(NO₃)₂ $\frac{2molKCl}{1molPb(NO_{3})_2}$ * $\frac{74.55g}{1molKCl}$ = 3.86 g KCl

3.86 g of KCl would react, so the amount remaining would be:

5.76 - 3.86 = 1.9 g KCl

8 0

3 years ago