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:
Metamorphic
Explanation:
Because it describes as a rock framed from other rocks by the action of heat and pressure.
Answer: The correct option is (c). The total pressure doubles.
Solution:
Initially, only 4 moles of oxygen gas were present in the flask.
(
) ( according to Dalton's law of partial pressure)
....(1)
= Total pressure when only oxygen gas was present.
Final total pressure when 4 moles of helium gas were added:
partial pressure of oxygen in the mixture :
Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.
= Total pressure of the mixture.
from (1)
On rearranging, we get:
The new total pressure will be twice of initial total pressure.
Magnesium in hydrochloric acid forms H2 (hydrogen) gas according to the balanced chemical equation:
Mg + 2HCl→H2 + MgCl2
The number of moles of gas lost is 0.0213 mol. It can be solved with the help of Ideal gas law.
<h3>What is Ideal law ?</h3>
According to this law, "the volume of a given amount of gas is directly proportional to the number on moles of gas, directly proportional to the temperature and inversely proportional to the pressure. i.e.
PV = nRT.
Where,
- p = pressure
- V = volume (1.75 L = 1.75 x 10⁻³ m³)
- T = absolute temperature
- n = number of moles
- R = gas constant, 8.314 J*(mol-K)
Therefore, the number of moles is
n = PV / RT
State 1 :
- T₁ = (25⁰ C = 25+273 = 298 K)
- p₁ = 225 kPa = 225 x 10³ N/m²
State 2 :
- T₂ = 10 C = 283 K
- p₂ = 185 kPa = 185 x 10³ N/m²
The loss in moles of gas from state 1 to state 2 is
Δn = V/R (P₁/T₁ - P₂/T₂ )
V/R = (1.75 x 10⁻³ m³)/(8.314 (N-m)/(mol-K) = 2.1049 x 10⁻⁴ (mol-m²-K)/N
p₁/T₁ = (225 x 10³)/298 = 755.0336 N/(m²-K)
p₂/T₂ = (185 x 10³)/283 = 653.7102 N/(m²-K)
Therefore,
Δn = (2.1049 x 10⁻⁴ (mol-m²-K)/N)*(755.0336 - 653.7102 N/(m²-K))
= 0.0213 mol
Hence, The number of moles of gas lost is 0.0213 mol.
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