The main <span>hazard </span>is the Radiation and the Gamma rays that are dispersed
Answer:
Explanation:
Let A₀ = the original amount of ⁵⁵Co
.
The amount remaining after one half-life is ½A₀.
After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.
After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀.
The general formula for the amount remaining is:
A =A₀(½)ⁿ
where n is the number of half-lives
n = t/t_½
Data:
A = 1.90 ng
t = 45 h
t_½ = 18.0 h
Calculation:
(a) Calculate n
n = 45/18.0 = 2.5
(b) Calculate A
1.90 = A₀ × (½)^2.5
1.90 = A₀ × 0.178
A₀ = 1.90/0.178 = 10.7 ng
The original mass of ⁵⁵Co was .
Answer:
Option C = object B by 1 gram per cubic cm.
Explanation:
Given data:
Mass of object A = 12 g
Volume of object A = 8 cm³
Mass of object B = 20 g
Volume of object B = 8 cm³
Densities = ?
Solution:
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Symbol:
The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.
Density of object A:
d = m/v
d = 12 g/ 8 cm³
d = 1.5 g/cm³
Density of object B:
d = m/v
d = 20 g/ 8 cm³
d = 2.5 g/cm³
object b has high density.
All of the acid molecules in beaker 1 dissociate fully and exist as and ions. As a result, beaker 1 represents a strong acid solution. The majority of the molecules in beaker 2 are undissociated.
1.00mol is the correct answer
