Given from the combined gas law that PV/T = k, there, therefore, the number of moles of the gas is constant.
According to the combined gas law, the product of the pressure and volume divided by the temperature is a constant. That is; PV/T = k
Looking at this statement we can see that the implied assumption in this law is that the number of moles of the gas is constant.
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Answer:
number of moles
Explanation:
I believe this is the right one :)
2.0 moles are represented by 118g of cobalt.
A mole is defined as 6.02214076 ×of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
Moles = 2.0023757 = 2.0moles
Hence, 2.0 moles are represented by 118g of cobalt.
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Answer:
The box will experience acceleration
Explanation:
The box experiences unbalanced forces and will, therefore, accelerate in other words, the force applied on it from the right is greater than that exerted on it from the left.
Answer:
tendons and ligaments
P1*V1 = P2*V2
760 mm Hg *30 L = 350 mm Hg * V2
V2 = 22800 mm Hg*L/350 mm Hg
V2 = about 65.14 L
2. A gas occupies 500 mL at 1 atm. What is its pressure if the volume becomes 1500 mL?
P1*V1 = P2*V2
1 atm*500 mL = P2*1500 mL
500 atm*mL/1500 mL = P2
P2 = 1/3 atm
3. If a gas occupies 250 mL at 25°C, what will its volume be at 57°C?
25°C + 273 = 298°K; 57°C + 273 = 330°K
V1/T1 = V2/T2
250 mL/298°K = V2/330°K
0.84 mL/k*330°K = V2
V2 = about 276.85 mL
4. What was the original temperature of a 2 L sample of gas if it initially had the volume 5 L at 27°C?
27°C + 273 = 298°K
V1/T1 = V2/T2
5 L/298°K = 2 L/T2
2 L/(0.017 L/K)= T2
T2 = 119.2°K = -153.8°C
5. At what pressure would 100 mL of a gas be found if it began at 500 mL at 2.5 atm?
P1*V1 = P2*V2
2.5 atm*500 mL = P2*100 mL
125 atm*mL/100 mL = P2
P2 = 1.25 atm
6. What is the volume of a gas at 275 mm Hg if it occupied 750 mL at 1000 mm Hg?
P1*V1 = P2*V2
1000 mm Hg*750 mL = 275 mm Hg*V2
750000 mm Hg*mL/275 mm Hg = V2
V2 = about 2727. 27 mL
7. A gas occupies 60 L at 250°K and 1 atm. What is its volume at 250°C and 1.5 atm?
250°C + 273 = 523°K
P1*V1/T1 = P2*V2/T2
1 atm*60 L/250°K = 1.5 atm*V2/523°K
0.24 (atm*L/K)*523°K = 1.5 atm*V2
125.52 (atm*L)/1.5 atm = V2
V2 = 83.68 L
8. What is the final temperature in Celsius of a gas occupying 3.25 L at 750 mm Hg if it occupied 2250 mL at 37°C and 600 mm Hg?
37°C + 273 = 310°K; 2250 mL = 2.25 L
P1*V1/T1 = P2*V2/T2
600 mm Hg*2.25 L/310°K = 750 mm Hg*3.25 L/T2
2437.35 mm Hg*L/ (4.35 mm Hg*L/K) = T2
T2 = 559.69°K
T2 = about 286.69°C