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Oxana [17]
3 years ago
12

What is one abiotic factor that all biotic factors need to survive??

Chemistry
1 answer:
labwork [276]3 years ago
5 0

Answer:

there r mainly three abiotic factors where all biotic factors need to thrust and survive,

They're i) Air

            ii) Water

           iii) Land

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4 years ago
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If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

4 0
3 years ago
Consider two electrochemical reactions: A and B. Reaction A results in the transfer of 2 mol of electrons per mole of reactant a
soldier1979 [14.2K]

Answer:

See explaination

Explanation:

The rate of the electrochemical reaction per unit area is given by

r=\frac{i}{nFA}

where r is the Reaction Rate (in mol s-1 cm-2)

i is the Current Generated (in A)

n is the Number of moles of electrons per mole of reactant

F is Faraday's Constant= 96,500 C mol-1

A is the Electrode Area (in cm2)

(Unit of current should be expressed in terms of Columb per second, C s-1 to obtain reaction rate in mol s-1 cm-2)

Net Reaction Rate for Reaction A:

Given: Number of moles of A, n=2 moles

Current generated, i=5 A=5 C s-1

Electrode Area, A= 2cm2

Thus reaction rate is

\mathbf{r=\frac{5 C s^{-1}}{2*96500 C mol^{-1} *2 cm^{2}}= 1.3*10^{-5} mol s^{-1} cm^{-2}}

Net Reaction Rate for Reaction B:

Given: Number of moles of B, n=3 moles

Current generated, i=15 A=15 C s-1

Electrode Area, A= 5cm2

Thus reaction rate is

\mathbf{r=\frac{15 C s^{-1}}{3*96500 C mol^{-1}*5cm^{2}}= 1.036*10^{-5} mol s^{-1} cm^{-2}}

Reaction A has higher net reaction rate.

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It occurs when the outermost electrons are permanently transferred to another atom.

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