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sesenic [268]
3 years ago
6

What developments occurred from the Jurassic period to the Cretaceous period?

Biology
2 answers:
yKpoI14uk [10]3 years ago
3 0
<span>Dinosaurs were common in the Jurassic period, but during the Cretaceous period they went extinct. Pangaea had fully split by the Cretaceous period and angiosperms had developed. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>
kaheart [24]3 years ago
3 0

Answer:

A) Dinosaurs became extinct

Explanation:

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3 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}

6 0
2 years ago
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