C) gas particles move rapidly but they collide with each other as well as on the wall of the container
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:
0.24 L of oxygen gas will be required to produce 0.160 L of nitrogen gas.
Explanation:
From the reaction, 3 moles of oxygen react with ammonia to produce 2 moles of nitrogen gas
So, 3 moles of 02 = 2 moles of N2
Since I mole of a gas occupies 22.4dm^3 or 22,4 L of the gas
3 * 22,4L of O2 produces 2 * 22.4 L of N2
67.2 L of O2 produces 44.8 L of N2
( 67.2 * 0.160 / 44.8 ) L of O2 will produce 0.160L of N2
10.752/ 44,8 L = 0.24 L of O2 will produce 0.160 L of N2
0.24 L of oxygen gas will be required to produce 0.160 L of nitrogen gas.
The answer to your question would be A. hope this helps...
