<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>
Answer:
the us and the soviet union
Explanation:
Answer:
your simpal answer is 177.32
Answer:
At equilibrium, reactants predominate.
Explanation:
For every reaction, the equilibrium constant is defined as the ratio between the concentration of products and reactants. Thus, for the reaction N2 (g) + O2 (g) ⇌ 2NO the expression of its equilibrium constant is:
![Keq = \frac{[NO]^{2}}{[O_{2} ][N_{2}]}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BO_%7B2%7D%20%5D%5BN_%7B2%7D%5D%7D)
Since the equilibrium constant is Keq = 4.20x10-31 the concentration of reactants O2 and N2 must be much higher than products to obtain such a small number as 4.20x10-31 at the equilibrium. Hence, at equilibrium reactants predominate.
