1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
finlep [7]
3 years ago
12

The average density of a carbon-fiber-epoxy composite is 1.615 g/cm3. the density of the epoxy resin is 1.21 g/cm3 and that of t

he carbon fibers is 1.74 g/cm3. (a) what is the volume percentage of carbon fibers in the composite? (b) what are the weight percentages of epoxy resin and carbon fibers in the composite?
Chemistry
1 answer:
ra1l [238]3 years ago
4 0
The composite material is composed of carbon fiber and epoxy resins. Now, density is an intensive unit. So, to approach this problem, let's assume there is 1 gram of composite material. Thus, mass carbon + mass epoxy = 1 g.

Volume of composite material = 1 g / 1.615 g/cm³ = 0.619 cm³
Volume of carbon fibers = x g / 1.74 g/cm³ 
Volume of epoxy resin = (1 - x) g / 1.21 g/cm³ 

a.) V of composite = V of carbon fibers + V of epoxy resin
0.619 = x/1.74 + (1-x)/1.21
Solve for x,
x = 0.824 g carbon fibers
1-x = 0.176 g epoxy resins

Vol % of carbon fibers = [(0.824/1.74) ÷ 0.619]*100 =<em> 76.5%</em>

b.) Weight % of epoxy = 0.176 g epoxy/1 g composite * 100 = <em>17.6%</em>
     Weight % of carbon fibers = 0.824 g carbon/1 g composite * 100 = <em>82.4%</em>
You might be interested in
exactly 5.00 L of air at -50.0 °C is warmed to 100.0 °C. What is the new volume if the pressure remains constant?
ollegr [7]

Answer:

V₂ = 8.36 L

Explanation:

Given data:

Initial volume of gas = 5.00 L

Initial temperature = -50.0°C (-50 + 273 = 223 k)

Final temperature = 100°C (100+273 = 373 k)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 5.00 L × 373 K / 223 K

V₂ = 1865 L.K / 223 K

V₂ = 8.36 L

3 0
3 years ago
What is one of the names for a column in the periodic table?
zheka24 [161]

Answer:

row, period, group

Explanation:

8 0
3 years ago
Read 2 more answers
On a class trip to the coast, you and your lab partner find a rock in the spot marked in red on the diagram. You show your teach
NNADVOKAT [17]
Cause it’s is the same
8 0
3 years ago
Read 2 more answers
Which of the following is not a physical change?
eimsori [14]
Combustion is not a physical change instead it's a chemical change because when we burn things we cannot regain it's original form.
6 0
3 years ago
What mass (in g) of potassium chlorate is required to supply the proper amount of oxygen needed to burn 117.3 g of methane
Brilliant_brown [7]

The mass (in g) of potassium chlorate required to supply the proper amount of oxygen needed to burn 117.3 g of methane is 1196.82 g

<h3>Combustion of methane</h3><h3 />

Methane burns in oxygen to produce carbon (iv) oxide and water according to the equation of the reaction below:

CH₄ + 2O₂  ----> CO₂ + 2H₂O

1 mole of methane requires 2 moles of oxygen for complete combustion

1 mole of methane has a mass of 16 g

moles of methane in 117.3 g = 117.3/16 = 7.33 moles of methane

7.33 moles of methane will require 2 * 7.33 moles of oxygen

7.33 moles of methane will require 14.66 moles of oxygen

<h3>Decomposition of potassium chlorate </h3>

The decomposition of potassium chlorate produces oxygen

The equation of the reaction is given below:

  • 2KClO3 → 2KCl + 3O2.

2 moles of potassium chlorate produces 3 moles of oxygen

14.66 moles of oxygen will be produced by 14.66 * 2/3  moles of potassium chlorate

14.66 moles of oxygen will be produced by 9.77 moles of potassium chlorate

1 mole of  potassium chlorate has a mass of 122.5

9.77 moles of potassium chlorate has a mass of 1196.82 g

Therefore, the mass (in g) of potassium chlorate required to supply the proper amount of oxygen needed to burn 117.3 g of methane is 1196.82 g

Learn more about mass and molar mass at: brainly.com/question/15476873

7 0
3 years ago
Other questions:
  • Remembering that SN2 reactions go with 100 % inversion of configuration, while SN1 reactions lead to racemization, explain why t
    13·1 answer
  • The molality of a solution containing 8.1 moles of solute in 4,847 g of solvent is _____. 1.7m 598m 39m 0.17m
    6·2 answers
  • Which change occurs when an atom in excited state returns to ground state? 1 energy is emitted 2energy is absorbed 3 the number
    12·2 answers
  • Name an element with electrons in the f subshell
    10·1 answer
  • Electrons involved in bonding between atoms are __________________
    14·1 answer
  • The element found naturally in the earth that has the heights atomic number is
    12·1 answer
  • A 2.52 L volume of hydrogen measured at −155 °C is warmed to 100°C. Calculate the volume (in L) of the gas at the higher tempera
    6·1 answer
  • Consider the energy diagram below.
    8·1 answer
  • A thorium-232 nucleus decays by a series of alpha and beta emissions until it reaches astatine-212. how many alpha emissions and
    15·1 answer
  • Can anyone help me in doing this question?
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!