Answer:
I know you have been waiting awhile for this question to be answered :)
Stoichiometry is used in industry quite often to determine the amount of materials required to produce the desired amount of products in a given useful equation. Each one of these products requires stoichiometry. There would be no products from these industries without chemical stoichiometry.
Explanation:
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Sorry you had to wait so long :(
Answer:
1.25 gram of cesium-137 will remain.
Explanation:
Given data:
Half life of cesium-137 = 30 year
Mass of cesium-137 = 5.0 g
Mass remain after 60 years = ?
Solution:
Number of half lives passed = Time elapsed / half life
Number of half lives passed = 60 year / 30 year
Number of half lives passed = 2
At time zero = 5.0 g
At first half life = 5.0 g/2 = 2.5 g
At 2nd half life = 2.5 g/ 2 = 1.25 g
Thus. 1.25 gram of cesium-137 will remain.
Answer:
the nucleus is the center of the atom, made up of protons and neutrons, without the nucleus you'd just have a bunch of electrons floating around; the nucleus is positively charged
protons are the positively charged particles that sit within the nucleus
neutrons are particles of no charge that sit within the nucleus, and because they have no charge, they do not cancel out the positive charge of the protons, making the nucleus positive
electrons are negatively charged particles that float around the nucleus in an area known as the electron cloud, they orbit around the nucleus because they are attracted to the positive charge of the nucleus (caused by the protons), with charges, opposites attract
Explanation:
Answer:
54
Barium atomic number is 56, this means that it has 56 protons in its nucleus and put it as a period 6 element. an uncharged barium atom would have 56 electron also. However as a group 2 element barium easily loses 2 electron to form the +2 cation. the ions has 56-2=54
Answer:
See explanation
Explanation:
In the Rutherford experiment, alpha particles were directed at the same spot on a thin gold foil.
As the alpha particles hit the foil, most of the alpha particles went through the foil. In Rutherford's interpretation, most of the particles went through because the atom consisted largely of empty space.
However, some of the alpha particles were deflected through large angles, in Rutherford's interpretation, the deflected alpha particles had hit the dense positive core of the atom which he called the nucleus.
This accounted for their scattering through large angles throughout the foil in all directions.
