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>
<em></em>
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>
<em></em>
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:
The answer to your question is 25 grams
Explanation:
Data
half-life = 5730 years
sample = 200 g
after 3 half-lives
Process
Calculate the amount of sample after one, two and three half-lives.
After each half-life, that of sample is half the previous amount.
Number of half-lives Amount of sample
0 200 g
1 100 g
2 50 g
3 25 g
Ho123 right I’m sorry of I’m wrong
Answer:
Active transport
Explanation:
Sodium-potassium pumps are examples of Active type of cellular transport. Sodium potassium pump exchanges sodium ions from potassium ions through the plasma membrane of animal cells.
Whereas Active transport can be defined as movement of ions and molecules across a cell membrane to the region of higher concentration with the help of enzymes and energy.