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Law Incorporation [45]
3 years ago
11

Research designed to answer a specific question or to solve a practical problem is the goal of _____.

Chemistry
1 answer:
kramer3 years ago
3 0

Answer:

Applied Chemistry.

Have a good day!

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Latent heat is used to: 1. form chemical bonds. 2. change molecular structure. 3. change states of matter. 4. change the tempera
likoan [24]

Answer : The correct option is, (3) change states of matter.

Explanation :

Latent heat : It is defined as the heat required to convert the solid into liquid or vapor and a liquid into a vapor without changing the temperature.

There are two types of latent heat.

(1) Latent heat of fusion

(2) Latent heat of vaporization

Latent heat of fusion : It is defined as the amount of heat energy released or absorbed when the solid converted to liquid at atmospheric pressure at its melting point.

Latent heat of vaporization : It is defined as the amount of heat energy released or absorbed when the liquid converted to vapor at atmospheric pressure at its boiling point.

Hence, latent heat is used to change states of matter.

7 0
3 years ago
How many moles are present in 356.4 g of NiBr3
lozanna [386]

1.194 mol

(remember to use sig figs!)

8 0
3 years ago
How do the fusion reaction in the sun compare to the fusion occurring in large atars and supernovas
r-ruslan [8.4K]
The fusion reaction in the sun is a combination of hydrogen atoms fusing to create helium. The fusion reaction in larger stars involve much heavier elements like oxygen and iron. In supernovas, often elements like gold are produced
6 0
3 years ago
Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe
Illusion [34]

Answer:

a. Kp=1.4

P_{A}=0.2215 atm

P_{B}=0.556 atm

b.Kp=2.0 * 10^-4

P_{A}=0.495atm

P_{B}=0.00995 atm

c.Kp=2.0 * 10^5

P_{A}=5*10^{-6}atm

P_{B}=0.9999 atm

Explanation:

For the reaction  

A(g)⇌2B(g)

Kp is defined as:

Kp=\frac{(P_{B})^{2}}{P_{A}}

The conditions in the system are:

          A                    B

initial   0                1 atm

equilibrium x       1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.    

Replacing these values in the expression for Kp we get:

Kp=\frac{(1-2x)^{2}}{x}

Working with this equation:

x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a} (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128

x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215

With x1 we get a partial pressure of:

P_{A}=1.128 atm

P_{B}=1-2*1.128 = -1.256 atm

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

P_{A}=0.2215 atm

P_{B}=1-2*0.2215 = 0.556 atm

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

P_{A}=0.505atm

P_{B}=1-2*0.505 = -0.01005 atm

Not sense.

With x2

P_{A}=0.495atm

P_{B}=1-2*0.495 = 0.00995 atm

X2 is the solution for this equation.  

Kp=2*10^5

X1=50001

X2=5*10^{-6}

With x1

P_{A}=50001atm

P_{B}=1-2*50001=-100001atm

Not sense.

With x2

P_{A}= 5*10^{-6}atm

P_{B}=1-2*5*10^{-6}= 0.9999 atm

X2 is the solution for this equation.  

8 0
3 years ago
Using the Polymers: A Property Database provide the following
Novosadov [1.4K]

Polyethene is a polymer composed of repeating units of the monomer ethene.

The properties of polyethene are as follows:

  • density- ranges 0.857 g/cm3 to 0.975 g/cm3.
  • specific heat capacity is 1.9 kJ/kg.
  • melting temperature is approximately 110 °C.

<h3>What are polymers?</h3>

Polymers are large macromolecules consisting of long repeating chains of smaller molecules known as monomers.

An example of a polymer is polyethene composed of repeating units of the monomer ethene.

The density of polyethylene ranges 0.857 g/cm3 to 0.975 g/cm3.

The specific heat capacity of polyethene is 1.9 kJ/kg.

The melting temperature of polyethene is approximately 110 °C.

Learn more about polyethene at: brainly.com/question/165779

4 0
2 years ago
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