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just olya [345]
3 years ago
15

Describe a Mesosaurus. ​

Chemistry
1 answer:
ElenaW [278]3 years ago
5 0

Mesosaurus lived in freshwater lakes and ponds. Elongated and slim, it measured about 3.3 feet long. The skull and tail were both long and narrow, and the animal probably undulated through the water as it fed on small crustaceans and other prey with its jaws, which were full of long, thin, pointed teeth.

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A cylinder at left with balls evenly spaced throughout the cylinder has an arrow leading to a cylinder at right cylinder with ba
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gas to liquid

Explanation:

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Cylinder to the left is filled with gases

Cylinder to the right is made up of liquid.

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learn more:

Phase change brainly.com/question/1875234

#learnwithBrainly

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A person would weigh more on Neptune than on planet earth because the gravitational field strength is stronger on Neptune.

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3 years ago
What volume of 0.10 M NaOH is required to neutralize 30 mL of 0.10 M HCl?
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30 mL. The two compounds need to balance each other out.
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Calculate the pH of a solution that is 0.235M benzoic acid and 0.130M sodium benzoate, a salt whose anion is the conjugate base
lesantik [10]

Hello!

We have the following data:

ps: we apply Ka in benzoic acid to the solution.

[acid] = 0.235 M (mol/L)

[salt] = 0.130 M (mol/L)

pKa (acetic acid buffer) =?

pH of a buffer =?

Let us first find pKa of benzoic acid, knowing that Ka (benzoic acid) = 6.20*10^{-5}

So:

pKa = - log\:(Ka)

pKa = - log\:(6.20*10^{-5})

pKa = 5 - log\:6.20

pKa = 5 - 0.79

\boxed{pKa = 4.21}

Now, using the abovementioned data for the pH formula of a buffer solution or (Henderson-Hasselbalch equation), we have:

pH = pKa + log\:\dfrac{[salt]}{[acid]}

pH = 4.21 + log\:\dfrac{0.130}{0.235}

pH = 4.21 + log\:0.55

pH = 4.21 + (-0.26)

pH = 4.21 - 0.26

\boxed{\boxed{pH = 3.95}}\end{array}}\qquad\checkmark

Note:. The pH <7, then we have an acidic solution.

I Hope this helps, greetings ... DexteR! =)

8 0
3 years ago
What is comis blast???​
Lina20 [59]

Answer:

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Explanation:

<em>HOPE</em><em> </em><em>IT</em><em> </em><em>HELPS</em><em> </em>

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