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2 years ago

When defining concentration during diffusion, what exactly are we describing?

Biology

1 answer:

Daniel [21]2 years ago

7 0

Answer:

The amount of a substance in relation to the total volume is the concentration. During diffusion, molecules are said to flow down their concentration gradient, flowing from an area of high concentration to an area of low concentration.

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2 years ago

Be sure to answer all parts.

Aleks04 [339]

Answer:

pH = 3.215

Explanation:

From the given information;

Using the equation for the dilution of a stock solution:

Since moles of C5H5N = moles of HNO3

Then:

$M_{C_5H_5N}\times V_{C_5H_5N}= M_{HNO_3}\times V_{HNO_3}$

$0.0750 M \times 2.65 L = 0.407 M \times V_{HNO_3}$

$V_{HNO_3}= \dfrac{0.0750 M \times 2.65 L}{0.407 M}$

$V_{HNO_3}=488.33 \ mL$

The reaction between C5H5NH and H2O is as follows:

$C_5H_5N^+H + H_2O$ ⇄ $C_5H_5N + H_3O^+$

$Molarity \ of \ C_5H_5N^+H = \dfrac{moles}{addition \ of\ the\ total\ volume}$

$Molarity \ of \ C_5H_5N^+H = \dfrac{0.0750 \ M \times 2.65 \ L}{2.65 \ L + 0.48833 \ L}$

$Molarity \ of \ C_5H_5N^+H = \dfrac{0.19875 \ ML}{3.13833 L }$

$= \ 0.06333\ M$

Now, the next step is to draw out the I.C.E table.

$C_5H_5N^+H + H_2O$ ⇄ $C_5H_5N + H_3O^+$

I 0.06333 0 0

C - x x x

E 0.06333 -x x x

$K_a = \dfrac{10^{-14}}{1.7 \times 10^{-19}} \\ \\ K_a = 5.8824 \times 10^{-6}$

$K_a = \dfrac{[C_5H_5N][H_3O^+] }{[C_5H_5N^+H] } \\ \\ 5.8824 \times 10^{-6} = \dfrac{x^2}{0.06333 - x}$

Assuming x < 0.06333

$x^2 = 5.8824 \times 10^{-6} \times 0.06333$

Then

$x^2 = 3.72532392 \times 10^{-7} \\ \\ x= \sqrt{3.72532392 \times 10^{-7}} \\ \\ x = 6.1035 \times 10^{-4} \ M$

$[H_3O^+] = x = 6.1035 \times 10^{-4} \ M \\ \\ pH = -log (6.1035 \times 10^{-4}) \\ \\ \mathbf{\\ pH = 3.215}$

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2 years ago

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When defining concentration during diffusion, what exactly are we describing?​

When defining concentration during diffusion, what exactly are we describing?