Answer:
is the activity, measured in of a 50 mg sample of 90-Sr.
Explanation:
Half life of the 90-Sr ,
Activity coefficient of the 90-Sr = 

Mass of 90-Sr = 50 mg = 0.050 g
Molecular mass of 90-Sr = 90 g/mol
Moles of 90-Sr =
Number of atom in 0.0005555 moles of 90-Sr:


1 year = 
Activity measured in atoms per seconds:
= Number of atoms × 


Answer:
Specific heat of substance = 0.897 J/g.°C
Explanation:
Given data:
Specific heat of substance = ?
Mass of substance = 25.0 g
Heat absorbed = 493.4 J
Initial temperature = 12.0 °C
Final temperature = 34°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 34°C -12.0°C
ΔT = 22°C
493.4 J = 25 g ×c× 22°C
493.4 J = 550 g.°C ×c
c = 493.4 J / 550 g.°C
c = 0.897 J/g.°C
Answer:
The compound which will precipitate first will be AgI with 
Explanation:
In order to precipitate a salt solution the ionic product of salt must exceed the solubility product.
Given:
Ksp of AgI = 
Ksp of PbI2= 
As Ksp of AgI is very low it will precipitate faster than lead iodide.
Now, higher the concentration of AgI solution taken faster its precipitation.
In the given choice the highest concentration solution of AgI is 
I think its the mirror because condenser helps in allowing the amount of light to pass.