Answer:
The second experiment (reversible path) does more work
Explanation:
Step 1:
A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C
<em>(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm</em>
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Irreversible path: w =-Pex*ΔV
⇒ with Pex = 1.00 atm
⇒ with ΔV = 1.20 L
W = -(1.00 atm) * 1.20 L
W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J
<em>(b) The gas is allowed to expand reversibly and isothermally to the same final volume.</em>
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W = -nRTln(Vfinal/Vinitial)
⇒ with n = the number of moles = 0.200
⇒ with R = gas constant = 8.3145 J/K*mol
⇒ with T = 298 Kelvin
⇒ with Vfinal/Vinitial = 2.40/1.20 = 2
W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)
W = -343.5 J
The second experiment (reversible path) does more work
Answer:
7.21 grams is the mass of methane
Explanation:
We may use the Ideal Gases Equation to solve this:
P. V = n. R. T
Let's determine the moles of Ar
18 g . 1 mol/ 39.9 g = 0.451 mol
In both situations, volume, temperature and pressure are the same so the moles of methane will also be the same as Argon's.
Let's convert the moles to mass of CH4.
0.451 mol . 16g/1mol = 7.21 grams
To get this it helps to know the electronegativity numbers of the elements but it isn't required. You just need to know that Fluorine is the most electronegative element and that the farther away from Fluorine you are on the periodic table, the less electronegative you get. The one exception to this rule is hydrogen with actually has an electronegativity of 2.1 while lithium has one of 1.0. Also the higher difference in electronegativity between two atoms the more polar the bond is.
Now to start the question. H-Br could be a contender since H has an electronegativity number of 2.1 and Br is relatively close to Fluorine so we'll put that one aside for now. H-Cl knocks out A because both bonds have H but one bond has Br and the other has Cl. Cl is closer to Fluorine than Br so answer B is the contender now. For answer C, I and Br are too close to have a higher electronegativity difference than H-Cl so that one isn't it. Finally for answer D, I is much closer to Cl than H is so the electronegativity difference is much less, making your answer B.
Answer:
Pro exercise con suffication
Explanation:
...
Answer:
It is a longitudinal wave.
Explanation:
Hope this helped.
