The changes in the energy law of conservation of energy is Potential energy is converted to kinetic energy. Kinetic energy is converted into potential energy.
<h3>What is the law of conservation of energy?</h3>
Law of conservation of energy says that energy can neither be created nor destroyed, it just transformed from one form to another.
The energies are kinetic, potential, mechanical, gravitational, electrical, etc.
Thus, the changes in the energy law of conservation of energy is Potential energy is converted to kinetic energy. Kinetic energy is converted into potential energy.
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Answer:
\left \{ {{y=206} \atop {x=82}}Pb \right.
Explanation:
isotopes are various forms of same elements with different atomic number but different mass number.
Radioactivity is the emission of rays or particles from an atom to produce a new nuclei. There are various forms of radioactive emissions which are
- Alpha particle emission \left \{ {{y=4} \atop {x=2}}He \right.
- Beta particle emission \left \{ {{y=0} \atop {x=-1}}e \right.
- gamma radiation \left \{ {{y=0} \atop {x=0}}γ \right.
in the problem the product formed after radiation was Pb-206. isotopes of lead include Pb-204, Pb-206, Pb-207, Pb-208. they all have atomic number 82. which means the radiation cannot be ∝ or β since both radiations will alter the atomic number of the parent nucleus.
Only gamma radiation with \left \{ {{y=0} \atop {x=0}}γ \right. will produce a Pb-206 of atomic number 82 and mass number 206 , since gamma ray have 0 mass and has 0 atomic number.equation is shown below
\left \{ {{y=206} \atop {x=82}}Pb\right ⇒ \left \{ {{y=206} \atop {x=82}}Pb\right + \left \{ {{y=0} \atop {x=0}}γ\right.
Thus the atomic symbol is \left \{ {{y=206} \atop {x=82}}Pb\right
Answer:
filtration is the process of using a filter to remove solids from liquids or gasses.
Example:
an example of this is tea.
Fine particles, ground level ozone, sulfur dioxide, nitrogen dioxide, lead
The average atomic mass of her sample is 114.54 amu
Let the 1st isotope be A
Let the 2nd isotope be B
From the question given above, the following data were obtained:
- Abundance of isotope A (A%) = 59.34%
- Mass of isotope A = 113.6459 amu
- Mass of isotope B = 115.8488 amu
- Abundance of isotope B (B%) = 100 – 59.34 = 40.66%
- Average atomic mass =?
The average atomic mass of the sample can be obtained as follow:
Thus, the average atomic mass of the sample is 114.54 amu
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