The earths gravity attracts the molecules and collects the most near the surface. They all have weight and therefore have more pressure at the surface, as well. As you go higher, the attraction becomes less and these molecules some times fly off into space. This layer of equilibrium has the least of weight or pressure.
The various pressures are measured by precision instruments called barometers or pressure sensors and expressed in inches of mercury or millibars. <span>Air has a weight too, although not very much, If you "pile" the air mile high, the bottom pressure is heavier because of all the air sitting on top of it, therefore the pressure decreases with altitude, because there is less air "piled up" </span>
<span>An analogy would be the same with water.</span>
Planes have these instruments that tells the crew the altitude above sea level they are at when flying.
Answer:
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Explanation:
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We need to see your problem/question you’re talking about to answer it
Answer:
No we can’t cover the earth with one mole of pennies
Explanation:
We can’t cover the earth with one mole of pennies due to the fact that making this possible will require the flattening and extra expansion before it covers the whole surface.
The extra expansion and flattening with result in the loss of intermolecular forces and won’t remain as a solid and form other states of matter which won’t be able to cover the surface.
Answer:
A. 85.6 g
= 0.0856 kg.
B. 0.00027 mol/g
= 0.27 mol/kg.
C. 8.39 %
Explanation:
Given:
Molar concentration = 0.25 M
Molar weight of sucrose = 342.296 g/mol
Density of solution = 1.02 g/mL
Mass of water = 934.4 g.
Density in g/l = 1.020 g/ml * 1000ml/1 l
= 1020 g/l
Mass of solution in 1 l of solution = 1020 g
Mass of solution = mass of solvent + mass of solute
Mass of sucrose = 1020 - 934.4
= 85.6 g of sucrose in 1 l of solution.
A.
Density of sucrose = mass/volume
= molar mass/molar concentration
= 342.296 * 0.25
= 85.6 g/l
Number of moles = mass/molar mass
= 85.6/342.296
= 0.25 mol
B.
Molality = number of moles of solute/mass of solvent
= 0.25/934.4
= 0.00027 mol/g
C.
% mass of sucrose = mass of sucrose/total mass of solution * 100
= 85.6/1020 * 100
= 8.39 %