<span>a) 7.9x10^9
b) 1.5x10^9
c) 3.9x10^4
To determine what percentage of an isotope remains after a given length of time, you can use the formula
p = 2^(-x)
where
p = percentage remaining
x = number of half lives expired.
The number of half lives expired is simply
x = t/h
where
x = number of half lives expired
t = time spent
h = length of half life.
So the overall formula becomes
p = 2^(-t/h)
And since we're starting with 1.1x10^10 atoms, we can simply multiply that by the percentage. So, the answers rounding to 2 significant figures are:
a) 1.1x10^10 * 2^(-5/10.5) = 1.1x10^10 * 0.718873349 = 7.9x10^9
b) 1.1x10^10 * 2^(-30/10.5) = 1.1x10^10 * 0.138011189 = 1.5x10^9
c) 1.1x10^10 * 2^(-190/10.5) = 1.1x10^10 * 3.57101x10^-6 = 3.9x10^4</span>
<span>Boyle's Law is k = PV so
Initial k = 13.0 L x 4.0 atm = 52 L atm
Final kf = 6.5 L x 8 atm = 52 L atm
The gas obeys Boyle's Law
The answer with two significant figures separated by a comma is k = 52, kf = 52.</span>
your answer is c. two atoms of oxygen.
Answer:
16 g/mol
Explanation:
In CO2, it means we have 1 mole of carbon and 2 moles of oxygen.
However, we want to find the molar mass of just a single mole of oxygen.
Now, from tables of values of elements in electronic configuration, the molar mass of oxygen is usually approximately 16 g/mol.
In essence the molar mass is simply the atomic mass in g/mol