<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>
X4O10
Let molar mass of X be y
molar mass = 4y + 10 x 16 = 4y+160
so, moles = 85.2 / (4y+160)
Moles of oxygen = 10 x [85.2 / (4y+160) ]
Mass of oxygen = 16 x 10 x [85.2 / (4y+160) ]
which is 48.0
so, 48 = 16 x 10 x [85.2 / (4y+160) ]
Solve the equation to get y.
y = 31
Answer:
Population of duck and frog will change with the change
Explanation:
The complete question is
Scientists are studying animals in a large lake area. In this lake area, both owls and raccoons eat ducks, and ducks eat frogs. The data shows that recently the size of the raccoon population decreased. How will the decrease in the raccoon population affect the other populations? Be sure to explain whether the owl population, the duck population, and the frog population will change, and why.
- Owl population will change
-
Duck population will change
-
Frog population will change
Solution
Raccoon eat duck and duck eat frog. Now if the population of Raccoon decreases then the number of predators of duck will decrease thereby increasing the population of duck.
The higher will be the number of ducks, the more frogs they will consume thereby decreasing the population of frogs
Hence both the population of duck and frog will change with the change
Answer:
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Hope this helps
Excerpt from textbook