This behavior helps the prairie dog from being prey upon by coyote.
Prairie dogs live in burrow and they normally do not move much away from their burrows. When they sense danger they will dash quickly back into their burrows. This behavioral adaptation protect them from predators.
Use inversenormal on a calculator and type in .1 , 70 , 10 . That percent is for A and will determine at what mark an A will be. Do the same for the rest of the grades but change the first argument in the calculation
I think it’s 1 Carbon, 4 hydrogens.
Taking into account the reaction stoichiometry, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
CO₂ + 4 H₄ → CH₄ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CO₂: 1 mole
- H₄: 4 moles
- CH₄: 1 mole
- H₂O: 2 moles
<h3>Moles of CH₄ formed</h3>
The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?

<u><em>moles of CH₄= 340.4 moles</em></u>
Then, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
Learn more about the reaction stoichiometry:
brainly.com/question/24741074
brainly.com/question/24653699
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Answer:
4.76
Explanation:
In this case, we have to start with the <u>buffer system</u>:

We have an acid (
) and a base (
). Therefore we can write the <u>henderson-hasselbach reaction</u>:
![pH~=~pKa+Log\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=pH~%3D~pKa%2BLog%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
If we want to calculate the pH, we have to <u>calculate the pKa</u>:

According to the problem, we have the <u>same concentration</u> for the acid and the base 0.1M. Therefore:
![[CH_3COO^-]=[CH_3COOH]](https://tex.z-dn.net/?f=%5BCH_3COO%5E-%5D%3D%5BCH_3COOH%5D)
If we divide:
![\frac{[CH_3COO^-]}{[CH_3COOH]}~=~1](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D~%3D~1)
If we do the Log of 1:

So:

With this in mind, the pH is 4.76.
I hope it helps!