This model is the simplest model.
110% sure. :)
A. The nervous system works without the aid of other systems in order to provide a sufficiently rapid response.
Answer:
3
I would like to understand the temperature at which a substance will vaporize when dissolved in a liquid. I have researched this online for hours, but haven't found a conclusive answer. Is it the boiling point of the dissolved substance? I'm attempting to find the temperature at which caffeine vaporizes when dissolved in water or other vegetable glycerin.
Explanation:
I hope this helps a little bit
<u>Answer:</u> The mass defect for the formation of phosphorus-31 is 0.27399
<u>Explanation:</u>
Mass defect is defined as the difference in the mass of an isotope and its mass number.
The equation used to calculate mass defect follows:
![\Delta m=[(n_p\times m_p)+(n_n\times m_n)]-M](https://tex.z-dn.net/?f=%5CDelta%20m%3D%5B%28n_p%5Ctimes%20m_p%29%2B%28n_n%5Ctimes%20m_n%29%5D-M)
where,
= number of protons
= mass of one proton
= number of neutrons
= mass of one neutron
M = mass number of element
We are given:
An isotope of phosphorus which is 
Number of protons = atomic number = 15
Number of neutrons = Mass number - atomic number = 31 - 15 = 16
Mass of proton = 1.00728 amu
Mass of neutron = 1.00866 amu
Mass number of phosphorus = 30.973765 amu
Putting values in above equation, we get:
![\Delta m=[(15\times 1.00728)+(16\times 1.00866)]-30.973765\\\\\Delta m=0.27399](https://tex.z-dn.net/?f=%5CDelta%20m%3D%5B%2815%5Ctimes%201.00728%29%2B%2816%5Ctimes%201.00866%29%5D-30.973765%5C%5C%5C%5C%5CDelta%20m%3D0.27399)
Hence, the mass defect for the formation of phosphorus-31 is 0.27399
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
