Well a question to ask would be if the mass of the material has changed significantly as that would determine that the substance is radioactive or if there have been any high readings found by a Geiger meter in certain period of time
hope that helps
Answer:
15.0 µm
Step-by-step explanation:
Density = mass/volume
D = m/V Multiply each side by V
DV = m Divide each side by D
V = m/D
Data:
m = 1.091 g
D = 7.28 g/cm³
l = 10.0 cm
w = 10.0 cm
Calculation:
<em>(a) Volume of foil
</em>
V = 1.091 g × (1 cm³/7.28 g)
= 0.1499 cm³
(b) <em>Thickness of foil
</em>
The foil is a rectangular solid.
V = lwh Divide each side by lw
h = V/(lw)
= 0.1499/(10 × 10)
= 1.50 × 10⁻³ cm Convert to millimetres
= 0.015 mm Convert to micrometres
= 15.0 µm
The foil is 15.0 µm thick.
We will use Arrehenius equation
lnK = lnA -( Ea / RT)
R = gas constant = 8.314 J / mol K
T = temperature = 25 C = 298 K
A = frequency factor
ln A = ln (1.5×10 ^11) = 25.73
Ea = activation energy = 56.9 kj/mol = 56900 J / mol
lnK = 25.73 - (56900 / 8.314 X 298) = 2.76
Taking antilog
K = 15.8