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.
You would use the formula for Boyle's Law:
(P1) (V1) = (P2) (V2)
(101.5) (2.0) = (P2?) (.75)
*P2 = 270kPa (You're allowed 2 significant figures)
P = Pressure
V = Volume
Answer:
Yes, you can compost lettuce and other salad leaves. Be careful! If the salad is covered in lots of dressing, the oils or fats in the dressing may attract rats or other animals to your compost heap.
That when water is boiled in a open beaker and it disappears that it evaporates into the air
Answer:
molar mass = 180.833 g/mol
Explanation:
- mass sln = mass solute + mass solvent
∴ solute: unknown molecular (nonelectrolyte)
∴ solvent: water
∴ mass solute = 17.5 g
∴ mass solvent = 100.0 g = 0.1 Kg
⇒ mass sln = 117.5 g
freezing point:
∴ ΔTc = -1.8 °C
∴ Kc H2O = 1.86 °C.Kg/mol
∴ m: molality (mol solute/Kg solvent)
⇒ m = ( - 1.8 °C)/( - 1.86 °C.Kg/mol)
⇒ m = 0.9677 mol solute/Kg solvent
- molar mass (Mw) [=] g/mol
∴ mol solute = ( m )×(Kg solvent)
⇒ mol solute = ( 0.9677 mol/Kg) × ( 0.100 Kg H2O )
⇒ mol solute = 0.09677 mol
⇒ Mw solute = ( 17.5 g ) / ( 0.09677 mol )
⇒ Mw solute = 180.833 g/mol
