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>
Answer:
The reagents are
.
Explanation:
1-Methylenecyclopentene is treated with HBr form 1-bromo-1-methylcyclopentane, which is treated with strong base ethoxide ion and forms 1-methylcyclopent-1-ene.
This alkene is treated with osmium tetraoxide in the presence of sodium bisulfite to form target product.
The chemical reaction is as follows.
Answer:
2.49*10⁻¹² mol
Explanation:
Use Avogadro's Number for this equation (6.022*10²³). Divide Avogrado's by the number of atoms you have to find moles. You are answer should be 2.49*10⁻¹² mol.
Answer:
The equilibrium concentration of CH₃OH is 0.28 M
Explanation:
For the reaction: CO (g) + 2H₂(g) ↔ CH₃OH(g)
The equilibrium constant (Keq) is given for the following expresion:
Keq=
=14.5
Where (CH3OH), (CO) and (H2) are the molar concentrations of each product or reactant.
We have:
(CH3OH)= ?
(CO)= 0.15 M
(H2)= 0.36 M
So, we only have to replace the concentrations in the equilibrium constant expression to obtain the missing concentration we need:
14.5= 
14.5 x (0.15 M) x
= (CH₃OH)
0.2818 M = (CH₃OH)
The average speed is 0.28