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inna [77]
3 years ago
14

1.What is adaptation and why is it important?

Chemistry
1 answer:
IrinaVladis [17]3 years ago
4 0

Answer:

All organisms need to adapt to their habitat to be able to survive. This means adapting to be able to survive the climatic conditions of the ecosystem, predators, and other species that compete for the same food and space.

Explanation:

You might be interested in
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).
Degger [83]
a) before addition of any KOH : 

when we use the Ka equation & Ka = 4 x 10^-8 : 

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

           = 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

       = 9.2 x 10^-5 M

when PH = -㏒[H+]

   PH = -㏒(9.2 x 10^-5)

        = 4  

b)After addition of 25 mL of KOH: this produces a buffer solution 

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]


first, we have to get moles of HCIO= molarity * volume

                                                           =0.21M * 0.05L

                                                           = 0.0105 moles

then, moles of KOH = molarity * volume 

                                  = 0.21 * 0.025

                                  =0.00525 moles 

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L =  0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

                                        = 0.00525 / 0.075

                                        =0.07 M

and molarity of KCIO = moles KCIO / total volume

                                    = 0.00525 / 0.075

                                    = 0.07 M

and when Ka = 4 x 10^-8 

∴Pka =-㏒Ka

         = -㏒(4 x 10^-8)

         = 7.4 

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4 

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume 

                                                        = 0.21 M * 0.05L

                                                        = 0.0105 moles

then moles KOH = molarity * volume
                            =  0.22 M* 0.035 L 

                            =0.0077 moles 

∴ moles of HCIO remaining = 0.0105 - 0.0077=  8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume 

                                      = 8 x 10^-5 / 0.085

                                      = 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

                                          = 0.0077M / 0.085L

                                          = 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

                                   = 0.0105mol / 0.1 L

                                   = 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

       = 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO] 

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

           = 3.8
∴PH = 14- POH

        =14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

 when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M 

V2 is the excess volume added  of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

           = -㏒0.02 

           = 1.7

∴PH = 14- POH

       = 14- 1.7 

      = 12.3 
8 0
3 years ago
Read 2 more answers
Consider the reaction. 2 Al ( s ) + Fe 2 O 3 ( s ) heat −−→ Al 2 O 3 ( s ) + 2 Fe ( l ) If 17.3 kg Al reacts with an excess of F
IrinaVladis [17]

Answer:

32.7 kilograms of aluminium oxide  will be produced.

Explanation:

2 Al ( s ) + Fe_2O_3 ( s ) +heat\rightarrow Al_2O_3 ( s ) + 2 Fe ( l )

Mass of aluminum = 17.3 kg = 17300 g (1 kg = 1000 g )

Moles of aluminium = \frac{17300 g}{27 g/mol}=640.74 mol

According to reaction, 2 moles of aluminum gives 1 mole of aluminum oxide,then 640.74 moles of aluminum will give:

\frac{1}{2}\times 640.74 mol= 320.37 mol of aluminum oxide

Mass of 320.37 moles of aluminum oxides:

320.37 mol × 102 g/mol = 32,677.74 g = 32.67774 kg ≈ 32.7 kg

32.7 kilograms of aluminium oxide  will be produced.

6 0
3 years ago
The table below shows the volume of two samples, A and B, when placed in three containers of different volumes.
deff fn [24]

Answer: B is a gas because its volume is less than the volume of the containers.

Explanation: hope this help bye

7 0
2 years ago
Read 2 more answers
The density of aluminum is 2.70 g/cm'. If the volume of a piece of aluminum is 90.4 cm', what is
weeeeeb [17]

Answer:

33.4814814815

Explanation:

90.4/2.70

6 0
3 years ago
Does gas have its own density
Leokris [45]

Answer:

The identity does not matter because the variables of Boyle's law do not identify the gas.

Explanation:

The ideal gas law confirms that 22.4 L equals 1 mol.

6 0
3 years ago
Read 2 more answers
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