It would be 4 meters per second. With this you'd only have to take 180, ans divide 45 from it to finally get your answer! I hope all is well, and you end up passing. (:
Thank you for posting your question here at brainly. The <span>mole fraction of co if the h2 mole fraction is 0.22 and the o2 mole fraction is 0.58 is 0.20, below is the solution:
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mole fraction CO + mole fraction H2 + mole fraction O2 = 1
mole fraction CO = 1 - ( 0.22 + 0.58)=0.20
Answer:
The answer is: 51.8 g (86% of serving size)
Explanation:
In order to solve the problem, we have to first determine the number of moles there are in 11.0 g of sucrose. Sucrose has a molecular weight of 342 g (we calculate this from the molar mass of the elements : 12 x 12 g/mol C + 22 x 1 g/mol H + 11 x 16 g/mol O). So, we divide the mass (11.0 g) into the molecular weight of sucrose:
11.0 g sucrose x 1 mol/342 g sucrose= 0.032 mol
We have 0.032 mol of sucrose in a serving of 60 g. But we need less moles (0.0278 mol):
0.032 mol ------------ 60 g serving
0.0278 mol------------ x= 0.0278 mol x 60 g serving/0.032 mol
x= 51.8 g
So, lesser than 1 serving of 60 g must be eaten to consume 0.0278 mol os sucrose. Exactly, 51.8 g (which stands for a 86% of the serving size).
273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit
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>
