Lets say foxes hunt rabbits. If their are too many foxes then they will compete for the rabbits. The rabbits get outnumbered by the amount of foxes so the foxes kill all the rabbits and the rabbits could eventually go extinct.
The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below
=% m/m =mass of the solute/mass of the solution x100
let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100
grams of glucose is therefore =2/100 = y/400
by cross multiplication
100y = 800
divide both side by 100
y= 8.0 grams
Answer: Farmers were paid to practice soil-conserving techniques like crop rotation and terracing
Explanation:
read about it here: https://www.pbs.org/wgbh/americanexperience/features/dust-bowl-surviving-dust-bowl/
Answer:
pH = 13.1
Explanation:
Hello there!
In this case, according to the given information, we can set up the following equation:

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

And the resulting concentration of KOH and OH ions as this is a strong base:
![[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E-%5D%3D%5Cfrac%7B0.00576mol%7D%7B0.012L%2B0.032L%7D%3D0.131M)
And the resulting pH is:

Regards!