Answer:
121Sb=57.2%
123Sb=42.8%
Explanation:
We are given that
Atomic mass of 121Sb=120.904 amu
Atomic mass of 123Sb=122.904 amu
Average atomic mass of antimony=121.760 amu
We have to find the percent of each of the isotope.
Let x be the percent of 121Sb and 1-x be the percent of 123Sb.
Using formula of average atomic weight
Average atomic weight=atomic weight of 121Sb
percentage abundance of isotope 121Sb+atomic weight of 123Sb
percentage abundance of isotope 123Sb
Substitute the values
Percentage of 121Sb=
57.2%
Abundance of isotope 123Sb=1-0.572=0.428
Percentage of isotope 123Sb=
42.8%
Explanation:
The ratio of the areas is the square of the ratio of the radii.
A/A = 3.16² = 9.99
The ratio of the volumes is the cube of the ratio of the radii.
V/V = 3.16³ = 31.6
Density = mass/volume
Density = 40g/20cm^3
Density = 2g/cm^3
Explanation:
1) Kicking a ball.
2) Capture the ball by hand.
3) Push a car.
Answer:
Explanation:
Given:
a = 3.10 m/s^2
vf = 12.4 m/s
vi = -6.2 m/s
t = (vf - vi)/a
= (12.4 + 6.2)/3.1
= 6 s
displacement = (vf - vi)*t
= (12.4 + 6.2) * 6
= 111.6 m.
