Answer:
the other stars are much farther away from Earth than our sun
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:
pH = 11.216.
Explanation:
Hello there!
In this case, according to the ionization of ammonia in aqueous solution:
We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:
![Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D%20%5C%5C%5C%5C1.80x10%5E%7B-5%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.150-x%7D)
However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:
Which is also:
![[OH^-]=1.643x10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.643x10%5E%7B-3%7DM)
Thereafter we can compute the pOH first:
Finally, the pH turns out:
Regards!
