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2 years ago

Humans have developed ways to increase the carrying capacity of their environment. True or false

Chemistry

1 answer:

gladu [14]2 years ago

7 0

Answer:

false.nowadays humans are destroying forest,extracted more minerals,making pollution. in my view.

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A liter is the same as a _____.

Nata [24]

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2 years ago

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omplete the relationship between ph and pka for each of the following atomic view pictures of a buffered solution.

Yakvenalex [24]

The relationship between pH and pKa of buffer solution in given atomic view:

In figure I pH= pKa ( since [HA] =[A-] )

In figure II pH > pKa ( since [A-] > [HA] )

In figure III pH < pKa ( since [A-] < [HA] )

The pH and pKa are related by the Henderson-Hasselbalch equation. It should not be used for concentrated solutions, extremely low pH acids, or extremely high pH bases because it is simply an approximation.

pH = pKa + log(conjugate base/weak acid).

pH equals pKa plus log ([A-] / [HA]).

pH is determined by dividing the weak acid concentration by the log of the conjugate base concentration and the pKa value.

About halfway to the equivalence point:

pH = pKa

It's important to note that this equation is familiar with the connection because it is sometimes written for the Ka value rather than the pKa value.

pKa = – log Ka

Hence, value of pH depend on relative concentration of [A-] and HA]

To know more about Ka.

brainly.com/question/16035742

#SPJ4

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9 months ago

Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the followi

marta [7]

Answer : The value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

Explanation :

The following equilibrium reactions are :

(1) $2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2$ $K_1=5.40\times 10^{-16}$

(2) $2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l)$ $K_2=1.06\times 10^{10}$

(3) $CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g)$ $K_3=2.68\times 10^{-9}$

The final equilibrium reaction is :

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$ $K_{goal}=?$

Now we have to calculate the value of $K_{goal}$ for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

$K_{goal}=\sqrt{K_1\times K_2\times K_3}$