Draw the following structures and name them :

Poly(lactic acid) (PLA) has received much recent attention because the lactic acid monomer [CH3CH(OH)COOH] from which it is made

Grace [21]

Answer:

The structure is shown in the figure.

Explanation:

There is condensation polymersiation in case of lactic acid with the removal of water molecule (dehydration) leading to formation of poly lactic acid.

The structure of PLA is shown in the figure.

3 0

4 years ago

You are asked to perform an experiment. When you perform the experiment you notice the vessel is warming. What type of reaction

Brut [27]

Exothermic type of reaction is taking place because this type of reaction generates heat energy.

<h3>What is exothermic reaction?</h3>

Exothermic reactions refer to those chemical reactions that produce heat energy and release that heat energy to the surrounding environment.

So we can conclude that exothermic type of reaction is taking place because this type of reaction generates heat energy.

Learn more about reaction here: brainly.com/question/26018275

3 0

2 years ago

Can someone help me name this hydrocarbon?

ser-zykov [4K]

2-ethyl-4,4 -dimethyl hex-1-ene.

3 0

3 years ago

How many grams of oxygen are required to react with 2.7 g of dihydrogen sulfide

KatRina [158]

Answer:

Mass = 3.84 g

Explanation:

Given data:

Mass of hydrogen sulfide = 2.7 g

Mass of oxygen required = ?

Solution:

Chemical equation:

2H₂S + 3O₂ → 2H₂O + 2SO₂

Number of moles of hydrogen sulfide:

Number of moles = mass/ molar mass

Number of moles = 2.7 g / 34 g/mol

Number of moles = 0.08 mol

Now we will compare the moles of hydrogen sulfide with oxygen.

H₂S : O₂

2 : 3

0.08 : 3/2 ×0.08 = 0.12 mol

Mass of oxygen;

Mass = number of moles × molar mass

Mass = 0.12 mol × 32 g/mol

Mass = 3.84 g

3 0

4 years ago

A chemist designs a galvanic cell that uses these two half-reactions:

Nonamiya [84]

Answer :

(a) Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$

(b) Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$

(c) $O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

(d) Yes, we have have enough information to calculate the cell voltage under standard conditions.

Explanation :

The half reaction will be:

Reaction at anode (oxidation) : $Fe^{2+}\rightarrow Fe^{3+}+e^-$ $E^0_{anode}=+0.771V$

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$ $E^0_{cathode}=+1.23V$

To balance the electrons we are multiplying oxidation reaction by 4 and then adding both the reaction, we get:

Part (a):

Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$ $E^0_{anode}=+0.771V$

Part (b):

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$ $E^0_{cathode}=+1.23V$

Part (c):

The balanced cell reaction will be,

$O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

Part (d):

Now we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

$E^o=(1.23V)-(0.771V)=+0.459V$

For a reaction to be spontaneous, the standard electrode potential must be positive.

So, we have have enough information to calculate the cell voltage under standard conditions.

4 0

3 years ago