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>
Non metals .. .. . . ............
Answer:
2994 kJ
Explanation:
When one mol of ethane (C₂H₆) is combusted, 1451 kJ of heat is exchanged.
First we convert 61.9 g of C₂H₆ into moles, using its molar mass:
- 61.9 g ÷ 30 g/mol = 2.06 mol C₂H₆
Finally we <u>calculate how much heat is exchanged by the combustion of 2.06 moles of C₂H₆</u>:
- 2.06 mol * 1451 kJ/mol = 2994 kJ
It would be acidic based indicator.
0-6 is acidic
7 is neutral
8-14 is alkaline
Answer:
c
Explanation:
A B and D are all fosil fuses c is not
