Answer;
-(2) An atom is mostly empty space.
Experiment
-Rutherford conducted the "gold foil" experiment where he shot alpha particles at a thin sheet of gold. The conclusion that can be drawn from these experiment is that an atom is mostly empty space.
-Rutherford found that a small percentage of the particles were deflected, while a majority passed through the sheet. This caused Rutherford to conclude that the mass of an atom was concentrated at its center, as the tiny, dense nucleus was causing the deflections.
Answer:
Lithospheric plates are regions of Earth's crust and upper mantle that are fractured into plates that move across a deeper plasticine mantle. Earth's crust is fractured into 13 major and approximately 20 total lithospheric plates.
Explanation:
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Answer:
- <em>He realized that some elements had not been discovered.</em>
Explanation:
Some scientists that tried to arrange the list of elements known before Mendeleev include Antoine Lavoisier, Johann Döbereiner, Alexandre Béguyer de Chancourtois, John Newlands, and Julius Lothar Meyer.
<em>Dimitri Mendeleev</em> was so succesful that he is recognized as the most important in such work.
Mendeleev by writing the properties of the elements on cards elaborated by him, and "playing" trying to order them, realized that, some properties regularly (periodically) repeated.
The elements were sorted in increasing atomic weight (which is not the actual order in the periodic table), but when an element did not meet the pattern discovered, he moved it to a position were its properties fitted.
The amazing creativity of Mendeleev led him to leave blanks for what he thought were places that should be occupied by elements yet undiscovered. More amazing is that he was able to predict the properties of some of those elements.
When years after some of the elements were discovered, the genius of Mendeleev was proven because the "new" elements had the properties predicted by him.
The vapor pressure is obtained as 23.47 torr.
<h3>What is the vapor pressure?</h3>
Given that; p = x1p°
p = vapor pressure of the solution
x1 = mole fraction of the solvent
p° = vapor pressure of the pure solvent
Δp = p°(1 - x1)
Δp =x2p°
Δp = vapor pressure lowering
x2 = mole fraction of the of the solute
Number of moles of glycerol = 32.5 g/92 g/mol = 0.35 moles
Number of moles of water = 500.0 g/18 g/mol = 27.8 moles
Total number of moles = 0.35 moles + 27.8 moles = 28.15 moles
Mole fraction of glycerol = 0.35 moles/28.15 moles = 0.012
Mole fraction of water = 27.8 moles/28.15 moles =0.99
Δp = 0.012 * 23.76 torr
Δp = 0.285 torr
p1 = p° - Δp
p1 = 23.76 torr - 0.285 torr
p1 = 23.47 torr
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