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otez555 [7]
2 years ago
8

Migration is _______. a. the movement of organisms from a native location to a foreign location b. the movement of organisms fro

m a foreign location to a native location c. the movement of organisms from their water supply to their food supply d. the seasonal movement of organisms between locations Please select the best answer from the choices provided A B C DMigration is _______. a. the movement of organisms from a native location to a foreign location b. the movement of organisms from a foreign location to a native location c. the movement of organisms from their water supply to their food supply d. the seasonal movement of organisms between locations Please select the best answer from the choices provided A B C D
Chemistry
2 answers:
MArishka [77]2 years ago
8 0

Answer:

D is the correct answer. the seasonal movement of organisms between locations

sergij07 [2.7K]2 years ago
4 0

Answer:

A

Explanation:

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You have 500.0 ml of a buffer solution containing 0.30 m acetic acid (ch3cooh) and 0.20 m sodium acetate (ch3coona). what will t
Nataly [62]
First, we should get moles acetic acid = molarity * volume

                                                                =0.3 M * 0.5 L

                                                                =  0.15 mol

then, we should get moles acetate = molarity * volume

                                                           = 0.2 M * 0.5L

                                                           = 0.1 mol

then, we have to get moles of OH- which added:

moles OH- = molarity  * volume

                   = 1 M    * 0.02L

                  = 0.02 mol

when the reaction equation is:


                 CH3COOH  +  OH-  → CH3COO-   +  H2O


moles acetic acid after adding OH- = (0.15-0.02) 
                                             
                                                            =  0.13M                                       

moles acetate after adding OH- =  (0.1 + 0.02)

                                                      =   0.12 M

Total volume = 0.5 L + 0.02 L= 0.52 L

∴[acetic acid] = moles acetic acid after adding OH- / total volume

                        = 0.13mol / 0.52L

                       = 0.25 M

and [acetate ) = 0.12 mol / 0.52L
 
                        = 0.23 M

by using H-H equation we can get PH:

PH = Pka + ㏒[salt/acid]

when we have Ka = 1.8 x 10^-5

∴Pka = -㏒Ka 

        = -㏒ 1.8 x 10^-5

       = 4.7

So by substitution:

∴ PH = 4.7 + ㏒[acetate/acetic acid]

         = 4.7 + ㏒(0.23/0.25)

        = 4.66
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