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Ainat [17]
3 years ago
11

What evidence would support Maria's claim that the fossilized remains are that of a bird, class Aves, instead of a turtle, class

Reptilia? A) the eggs B) a backbone C) a keeled sternum D) warm-bloodedness
Chemistry
2 answers:
lora16 [44]3 years ago
6 0

Your answer would be C.

Ulleksa [173]3 years ago
3 0

Answer:

The correct answer is option c, that is, keeled sternum.

Explanation:

In the given question, there is a need to access the skeleton or the fossilized remains of the species. The turtles possess bony, tough shells that safeguard their body. It exhibits a domed top (carapace) and a flat layer underneath the belly (plastron). The backbones of the turtles are amalgamated with the bones in their shells.  

While, the birds exhibit a lightweight skeleton, which is formed majorly of thin and hollow bones. If the skeleton possesses a breastbone, or the keeled sternum, then what they have found is most probably the nest of the bird.  

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The jet stream is one of the key winds that moves air masses. in which direction does the jet stream in the united state general
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The jet stream is one of the key winds that moves air masses at the direction from west to east it moves.

<h3>What are jet streams?</h3>

Jet streams are strong winds which are flow in the form of narrow bands in the upper part of the atmosphere. On the earth these winds are flows near the altitude of the tropopause. These winds carries weather systems and blows toward the colder northern air and due to the rotation of earth, its direction is from west to east.

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Lewis structure for CO2
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3 years ago
A student throws a 2kg ball at 4m/s. What is the ball’s momentum?
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4 0
3 years ago
Uranium has three common isotopes. If the abundance of 234U is 0.0054%, the abundance of 235U is 0.7204% and the abundance of 23
OlgaM077 [116]

Answer:

237.9 amu

Explanation:

Given data:

Abundance of U-234 = 0.0054 %

Abundance of  U-235 = 0.7204%

Abundance of  U-238 = 99.2742%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass  / 100

Average atomic mass = (0.0054×234)+(0.7204×235) +(99.2742×238)  /100

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Average atomic mass = 237.9 amu.

7 0
3 years ago
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