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3 years ago

Which best describes what is represented by images 1 and 2?

Chemistry

2 answers:

viva [34]3 years ago

8 0

<u>Answer: </u>The correct answer is image 1 shows a monomer, and Image 2 shows a polymer.

<u>Explanation:</u>

A monomer is defined as the molecule that react with the same molecule to form a large molecule known as polymer. These are the repeating units in a polymer.

A polymer is a large molecule which are formed when a large number of same type of molecules react together.

A macro-molecule is defined as a very large molecule such as proteins which are formed by the polymerization of monomers. They are known as natural polymers.

A synthetic polymer are defined as the polymers which are made by humans. They are known as artificial polymers.

In the images given, Image 1 represents a monomer known as ethylene and Image 2 is a polymer known as polyethylene.

Hence, the correct answer is image 1 shows a monomer, and Image 2 shows a polymer,

arlik [135]3 years ago

5 0

the answer is

Image 1 shows a monomer, and Image 2 shows a polymer.

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You wish to construct a buffer of pH=7.0. Which of the following weak acids (w/ corresponding conjugate base) would you select?

Alex17521 [72]

Answer:

C.) HOCl Ka=3.5x10^-8

Explanation:

In order to a construct a buffer of pH= 7.0 we need to find the pKa values of all the acids given below

we Know that

pKa= -log(Ka)

therefore

A) pKa of HClO2 = -log(1.2 x 10^-2)

=1.9208

B) similarly PKa of HF= -log(7.2 x 1 0^-4)= 2.7644

C) pKa of HOCl= -log(3.5 x 1 0^-8)= 7.45

D) pKa of HCN = -log(4 x 1 0^-10)= 9.3979

If we consider the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution

The weak acid for making the buffer must have a pKa value near to the desired pH of the weak acid.

So, near to value, pH=7.0. , the only option is HOCl whose pKa value is 7.45.

Hence, HOCl will be chosen for buffer construction.

3 0

3 years ago

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

3 years ago

Electrolytic cells use electricity to cause a nonspontaneous redox reaction to occur. An electrolytic cell is constructed using

Fittoniya [83]

Answer:

At anode - ${Br_2}_{(l)}$

At cathode - ${H_2}_{(g)}$

Explanation:

Electrolysis of NaBr:

Water will exist as:

$H_2O\rightleftharpoons H^++OH^-$

The salt, NaBr will dissociate as:

$NaBr\rightarrow Na^++Br^-$

At the anode, oxidation takes place, as shown below.

${Br^-}_{(aq)}\rightarrow {Br_2}_{(l)}+2e^-$

At the cathode, reduction takes place, as shown below.

$2{H^+}_{(aq)}+2e^-\rightarrow {H_2}_{(g)}$

4 0

3 years ago

How many carbon atoms are there in .500 mol of CO2?

AURORKA [14]

Answer: There are $3.011 \times 10^{23}$ atoms present in 0.500 mol of $CO_{2}$ .

Explanation:

According to the mole concept, there are $6.022 \times 10^{23}$ atoms present in 1 mole of a substance.

In a molecule of $CO_{2}$ there is only one carbon atom present. Therefore, number of carbon atoms present in 0.500 mol of $CO_{2}$ are as follows.

$1 \times 0.500 \times 6.022 \times 10^{23}\\= 3.011 \times 10^{23}$

Thus, we can conclude that there are $3.011 \times 10^{23}$ atoms present in 0.500 mol of $CO_{2}$ .

6 0

2 years ago

explain what happens to the energy stored in an item after it stops working or is no longer functional. Explain in depth

gulaghasi [49]

Answer:

it gets rust and could not run properly

6 0

3 years ago