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Elanso [62]
3 years ago
13

Which term best describes this reaction?

Chemistry
2 answers:
lesya692 [45]3 years ago
7 0
 Try the second or  third
Sonja [21]3 years ago
4 0

Answer:

The best term that describes the reaction is addition polymerization

Explanation:

Polymerization is a process which small molecules, monomers, undergo a chemical reaction to form polymer chains. The two general types are: addition polymerization and condensation polymerization.

Addition polymerization occurs when monomers reacts by the opening of the double bond to form a long carbon chain. The reaction continues until the reactive intermediate is destroyed in a termination reaction. This is a major process of producing polymers such as: poly(ethene) and poly(propene). It process has three stages which are: initiation, propagation, and termination.

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What is the difference in concentration between a pH of 7 and 12?
Ratling [72]

Answer:

The pH of a solution is simply a measure of the concentration of hydrogen ions,  

H

+

, which you'll often see referred to as hydronium cations,  

H

3

O

+

.

More specifically, the pH of the solution is calculated using the negative log base  

10

of the concentration of the hydronium cations.

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

pH

=

−

log

(

[

H

3

O

+

]

)

a

a

∣

∣

−−−−−−−−−−−−−−−−−−−−−−−−  

Now, we use the negative log base  

10

because the concentration of hydronium cations is usually significantly smaller than  

1

.

As you know, every increase in the value of a log function corresponds to one order of magnitude.

Explanation:

4 0
3 years ago
Read 2 more answers
List four observations that suggest chemical change is occurring
allsm [11]

Answer:

Examples of Chemical Changes

Burning wood.

Souring milk.

Mixing acid and base.

Digesting food.

Cooking an egg.

Heating sugar to form caramel.

Baking a cake.

Rusting of iron.

8 0
3 years ago
4. How many grams of ammonium carbonate are needed to decompose in order to produce
Thepotemich [5.8K]

Answer:

14.23g of (NH4)2CO3

Explanation:

We'll begin by writing the balanced equation for the reaction.

(NH4)2CO3 –> (NH4)2O + CO2

Next,, we shall determine the mass of (NH4)2CO3 that decomposed and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of (NH4)2CO3 = 2[14+(4x1)] + 12 + (16x3)

= 2[14 +4] + 12 + 48

= 2[18] + 60 = 96g/mol

Mass of (NH4)2CO3 from the balanced equation = 1 x 96 = 96g

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

Mass of CO2 from the balanced equation = 1 x 44 = 44g.

Summary:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Finally, we can determine the mass of (NH4)2CO3 that decomposed to produce 6.52g of CO2 as follow:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Therefore, Xg of (NH4)2CO3 will decompose to produce 6.52g of CO2 i.e

Xg of (NH4)2CO3 = (96 x 6.52)/44

Xg of (NH4)2CO3 = 14.23g

Therefore, 14.23g of (NH4)2CO3 is needed to produce 6.52g of CO2.

4 0
3 years ago
Atoms of the same element can have different forms; for example, carbon-12 and carbon-14.
cricket20 [7]
A carbon-12 atom has 6 protons (6P) and 6 neutrons (6N). But some types of carbon have more than six neutrons. We call forms of elements that have a different number of neutrons, isotopes. For example, carbon-14 is a radioactive isotope of carbon that has six protons and eight neutrons in its nucleus.
Hope that helps
5 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
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