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>
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1 half-life - 4.5 days
x half-lives - 27 days
x=(1*27)/4.5=6
6 <span>half-lives have elapsed after 27 days.</span>
Answer:
30.3 g
Explanation:
At STP, 1 mol of any gas will occupy 22.4 L.
With the information above in mind, we <u>calculate how many moles are there in 32.0 L</u>:
- 32.0 L ÷ 22.4 L/mol = 1.43 mol
Then we <u>calculate how many moles would there be in 16.6 L</u>:
- 16.6 L ÷ 22.4 L/mol = 0.741 mol
The <u>difference in moles is</u>:
- 1.43 mol - 0.741 mol = 0.689 mol
Finally we <u>convert 0.689 moles of CO₂ into grams</u>, using its <em>molar mass</em>:
- 0.689 mol * 44 g/mol = 30.3 g
Coal, oil, and natural gas are the 3 common fossil fuels
