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olga_2 [115]
2 years ago
7

Scientists have collected the following data regarding rock ages along a mid-ocean ridge.

Chemistry
1 answer:
ivanzaharov [21]2 years ago
4 0

The statement is supported by the data, is that the youngest rocks on the seafloor were at the mid-ocean ridge.

<h3>What is the evidence regarding the age of the rocks?</h3>

The data collected by scientists regarding rock ages along a mid-ocean ridge seafloor with mid-ocean ridge showing age of rock increasing the further from the ridge.

From the data, they found that the youngest rocks on the seafloor were at the mid-ocean ridge and the rocks get older with distance from the ridge crest.

Thus, the statement is supported by the data, is that the youngest rocks on the seafloor were at the mid-ocean ridge.

Learn more about age of rocks here: brainly.com/question/23546028

#SPJ1

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50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the
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a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

H_2NC_2H_5CO^-_2  +  H^+  ------>  H_3}^+NC_2H_5CO^-_2

1 mole of  alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL  of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via  the following process as shown below;

[H_3}^+NC_2H_5CO^-_2] = \frac{initial moles of alaninate}{total volume}

[H_3}^+NC_2H_5CO^-_2] = \frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}

[H_3}^+NC_2H_5CO^-_2] = \frac{8}{100mL}

[H_3}^+NC_2H_5CO^-_2] = 0.08 M

Alanine serves as an intermediary form, however the concentration of H^+ and the pH can be determined as follows;

[H^+] = \sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{  K_{a1}{[H_3}^+NC_2H_5CO^-_2]  } }

[H^+] = \sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})}  {(10^{-pK_{a1}})+(0.08)} }

[H^+] = \sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})}  {(10^{-2.344})+(0.08)} }

[H^+] =  7.63*10^{-7}M

pH = - log [H^+]

pH = -log[7.63*10^{-7}]

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

H_2NC_2H_5CO^-_2    +  H^+     ----->   H^+_3NC_2H_5CO^-_2

H^+_3NC_2H_5CO^-_2    +  H^+     ----->   H^+_3NC_2H_5CO_2H

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

[H^+_3NC_2H_5CO_2H] = \frac{initial moles of alaninate}{total volume}

[H^+_3NC_2H_5CO_2H] = \frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}

[H^+_3NC_2H_5CO_2H] = \frac{8}{150}

[H^+_3NC_2H_5CO_2H] = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

H^+_3NC_2H_5CO_2H        ⇄        H^+_3NC_2H_5CO^-_2    +  H^+

(0.053 - x)                                  x                             x

K_{a1} = \frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}

10^{-PK_{a1}} = \frac{x*x}{(0.053-x)}

10^{-2.344} =\frac{x^2}{(0.053-x)}

0.00453 = \frac{x^2}{(0.053-x)}

0.00453(0.053-x) =x^2

x^2+0.00453x-(2.4009*10^{-4})

Using quadratic equation formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

we have:

\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)} OR \frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}

= 0.0134                    OR                -0.0179

So; we go by the positive integer which says

x = 0.0134

So [H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M

pH = -log[H^+]

pH = -log[0.0134]

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

3 0
3 years ago
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