<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
<h2>~<u>Solution</u> :-</h2>
- Here, to find the atomic mass of element, we must;
We know that,
- 4.6 x $ \sf{10^{22}}$ atoms of an element weigh 13.8g.
Thus,
The atoms of $ \sf{ 6.02 \times 10^{13}}$ will weigh;
- Hence, the molar mass (atomic mass) will be <u>180.6 g.</u>
The halogens are five non-metallic elements found in group 17 of the periodic table. The term "halogen" means "salt-former" and compounds containing halogens are called "salts". All halogens have 7 electrons in their outer shells, giving them an oxidation number of -1.
B. Two significant figures
Your final answer can only have as many significant figures as the number that has the least amount of significant figures in it, which is (0.0065)
Answer:
Answer E.
For a collision to be completely elastic, there must be NO LOSS in kinetic energy.
We can go through each answer choice:
A. Since the ball rebounds at half the initial speed, there is a loss in kinetic energy. This is NOT an elastic collision.
B. A collision involving sticking is an example of a perfectly INELASTIC collision. This is NOT an elastic collision.
C. A reduced speed indicates that there is a loss of kinetic energy. This is NOT elastic.
D. The balls traveling at half the speed after the collision indicates a loss of kinetic energy, making this collision NOT elastic.
E. This collision indicates an exchange of velocities, characteristic of an elastic collision. We can prove this:
Let:
m = mass of each ball
v = velocity
We have the initial kinetic energy as:
KE = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2KE=21mv2+0=21mv2
And the final as:
KE = 0 + \frac{1}{2}mv^2 = \frac{1}{2}mv^2KE=0+21mv2=21mv2
