Given:
Half life(t^ 1/2) :30 years
A0( initial mass of the substance): 200 mg.
Now we know that
A= A0/ [2 ^ (t/√t)]
Where A is the mass that remains after t years.
A0 is the initial mass
t is the time
t^1/2 is the half life
Substituting the given values in the above equation we get
A= [200/ 2^(t/30) ] mg
Thus the mass remaining after t years is [200/ 2^(t/30) ] mg
Answer:
The molar mass of lysine using the ideal gas equation for this problem is 146.25 g/mole.
Explanation:
The ideal gas equation PV = nRT, was derived from the ABC laws (Avogadros, Boyles and Charles laws). We need to obtain the value for the number of moles n.
The parameters of this equation are:
P = 1.918 atm
V = 750.0mL = 0.75L
n = ?
R = 0.0821
T = 25 degree celcius = 25 + 273 = 298 degree kelvin.
From this formular, n = (PV)/(RT)
n = (1.918 X 0.75)/(0.0821 X 298 )
n = 0.0588
n, no of mole = mass/molar mass
0.0588 = 8.6/MM
MM = 8.6/0.0588
MM = 146.25g/mole.
8,002.5 would be written as
8.0025 x 10 and 3
because you move the decimal place 3 times to the left.
