Since 1.00 atm is equal to 760 mmHg, the height of the column of mercury which would create pressure equal to 1.00 atm is 760 mm or 76 cm
<h3>What are the units of pressure?</h3>
Pressure is defined as the force acting perpendicularly per unit area of a surface.
The units of pressure include;
- millimetres of mercury, mmHg
- aatmospheres, atm,
- pascals, Pa
- Newtons per square metre, N/m^2
1 atmosphere is equal to 760 mmHg.
Therefore, the height of the column of mercury which would create pressure equal to 1.00 atm is 760 mm or 76 cm
Learn more about pressure at: brainly.com/question/25934121
Answer:
earth land assembling together to form single large landmass
It would still have oceans but no atmospheric water in Earth if no icy debris had arrived.
A. It would still have oceans but no atmospheric water.
<u>Explanation:</u>
Seas characterize our home planet, covering most of the Earth's surface and driving the water cycle that commands our territory and climate. However, progressively significant still, the narrative of our seas wraps our home in a far bigger setting that ventures profound into the universe and spots us in a rich group of sea universes that range our nearby planetary group and past.
It would in any case have seas yet no air water on Earth if no frigid flotsam and jetsam had shown up. For a long time, it was accepted that the frosty moons were only that - solidified husks, strong to their center. However, lately that thought has steadily been supplanted by a fresher, additionally energizing worldview.
Answer:
The answer is most likely A. Definite proportions
Explanation:
The Law of Definite proportions states that a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.
Answer:
3. V = 0.2673 L
4. V = 2.4314 L
5. V = 0.262 L
6. V = 2.224 L
Explanation:
3. assuming ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V1 = 225 L
∴ T1 = 175 K
∴ P1 = 150 KPa = 1.48038 atm
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))
⇒ n = 0.043 mol
∴ T2 = 112 K
∴ P2 = P1 = 150 KPa = 1.48038 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)
⇒ V2 = 0.2673 L
4. gas is heated at a constant pressure
∴ T1 = 180 K
∴ P = 1 atm
∴ V1 = 44.8 L
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))
⇒ n = 0.3295 mol
∴ T2 = 90 K
⇒ V2 = RT2n/P
⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)
⇒ V2 = 2.4314 L
5. V1 = 200 L
∴ P1 = 50 KPa = 0.4935 atm
∴ T1 = 271 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))
⇒ n = 0.2251 mol
∴ P2 = 100 Kpa = 0.9869 atm
∴ T2 = 14 K
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)
⇒ V2 = 0.262 L
6.a) ∴ V1 = 24.6 L
∴ P1 = 10 atm
∴ T1 = 25°C = 298 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))
⇒ n = 0.0993 mol
∴ T2 = 273 K
∴ P2 = 101.3 KPa = 0.9997 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)
⇒ V2 = 2.224 L
