Please answer this question correctly

Draw the carboxylic acid produced from the acid hydrolysis of butyl acetate.

Nat2105 [25]

Answer:

Acetic or ethanoic acid.

Explanation:

Hello,

Butyl acetate's hydrolisis is shown below:

$CH_3COOCH_2CH_2CH_2CH_3+H_2O-->CH_3COOH+CH_3CH_2CH_2CH_2OH$

Whereas $CH_3COOCH_2CH_2CH_2CH_3$ is the butyl acetate, $CH_3CH_2CH_2CH_2OH$ butanol and $CH_3COOH$ the required produced acid, acetic or ethanoic acid (attached picture).

Best regards.

7 0

3 years ago

A 3.96x10^-24 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000-cm cuvet; a blank solution containi

viktelen [127]

Actual question from source:-

A 3.96x10-4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000 cm cuvette. A blank had an absorbance of 0.029. The absorbance of an unknown solution of compound A was 0.375. Find the concentration of A in the unknown.

Answer:

Molar absorptivity of compound A = $1502.53\ {Ms}^{-1}$

Explanation:

According to the Lambert's Beer law:-

$A=\epsilon l c$

Where, A is the absorbance

l is the path length

$\epsilon$ is the molar absorptivity

c is the concentration.

Given that:-

c = $3.96\times 10^{-4}\ M$

Path length = 1.000 cm

Absorbance observed = 0.624

Absorbance blank = 0.029

A = 0.624 - 0.029 = 0.595

So, applying the values in the Lambert Beer's law as shown below:-

$0.595=\epsilon\times 1.000\ cm\times 3.96\times 10^{-4}\ M$

$\epsilon=\frac{0.595}{3.96\times 10^{-4}}\ {Ms}^{-1}=1502.53\ {Ms}^{-1}$

<u>Molar absorptivity of compound A = $1502.53\ {Ms}^{-1}$ </u>

4 0

4 years ago

Okay im clueless and over thinking this question can someone please help me out

Crank

A cookie made from flour, eggs, sugar, butter, and chocolate chips is an example of a heterogeneous mixture.

Also, mixing the sugar flavoring in water, since you're mixing two different states of matter.

6 0

3 years ago

1.

fgiga [73]

Answer:

THE PRESSURE OF THE TIRE ON THE TRIP HOME AT THE ROAD SURFACE TEMPERATURE OF 32°C IS 160 kPa.

Explanation:

Initial Pressure = 75 kPa

Initial temperature = 15 °C

Final temperature = 32 °C

Final pressure = unknown

Using the combined equation of gases;

P1V1/T1 = P2V2/ T2

Since the tire will have the same volume of air in it showing that volume of constant both at the repair shop and on the road surface.

The relationship between pressure and temperature is used with constant volume.

P1/T1 = P2/ T2

75 kPa / 15 °C = P2 / 32 °C

P2 = 75 kPa * 32 °C / 15 °C

P2 = 2400 kPa °C / 15 °C

P2 = 160 kPa.

So therefore, the pressure of the tire on the trip home when the temperature of the road surface is 32°C is 160 kPa.

5 0

4 years ago

How many grams of CO2 will be produced when 8.50 g of methane react with 15.9 g of O2, according to the following reaction? CH4(

Vedmedyk [2.9K]

Taking into account the reaction stoichiometry, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CH₄: 1 mole
O₂: 2 moles
CO₂: 1 mole
H₂O: 2 moles

The molar mass of the compounds is:

CH₄: 16 g/mole
O₂: 32 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

CH₄: 1 mole ×16 g/mole= 16 grams
O₂: 2 moles ×32 g/mole= 64 grams
CO₂: 1 mole ×44 g/mole= 44 grams
H₂O: 2 moles ×18 g/mole=36 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 16 grams of CH₄ reacts with 64 grams of O₂, 8.50 grams of CH₄ reacts with how much mass of O₂?

$mass of O_{2} =\frac{8.50 grams of CH_{4}x64 grams of O_{2} }{16grams of CH_{4}}$

mass of O₂= 34 grams

But 34 grams of O₂ are not available, 15.9 grams are available. Since you have less mass than you need to react with 8.50 grams of CH₄, O₂ will be the limiting reagent.

<h3>Mass of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 64 grams of O₂ form 44 grams of CO₂, 15.9 grams of O₂ form how much mass of CO₂?

$mass of CO_{2} =\frac{15.9 grams of O_{2}x44 grams of CO_{2} }{64grams of O_{2}}$

<u><em>mass of CO₂= 10.93 grams</em></u>

Then, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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6 0

2 years ago