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skad [1K]
2 years ago
13

You are studying a large tropical reptile that has a high and relatively stable body temperature. How would you determine whethe

r this animal is an endotherm or an ectotherm?
Chemistry
1 answer:
Vesnalui [34]2 years ago
8 0

Explanation:

Endothermic animals are also known as warm-blooded, they have the capacity to regulate their body temperature independent of the environment. They have mechanisms to compensate if heat loss exceeds heat generation (shivers) Or if heat generation exceeds the heat loss (panting, sweating).

On the other hand, ectothermal animals are known as cold blooded organisms and depend on external sources, like sunlight, to regulate their body temperature, reptiles are ectothermals.

To determine if the animal of interest is endo or ectothermal you’ll have to consider that is a reptile, you’ll also observe that it consumes less food and finally it’ll have more difficulties to adapt to sudden temperature changes.

I hope you find this information useful and interesting! Good luck!

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A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research techni
torisob [31]

Answer:

a. pka = 3,73.

b. pkb = 10,27.

Explanation:

a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:

HX + H₂O ⇄ H₃O⁺ + X⁻

Where ka is defined as:

ka = \frac{[H_3O^+][X^-]}{[HX]}

In equilibrium, molar concentrations are:

[HX] = 0,089M - x

[H₃O⁺] = x

[X⁻] = x

pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:

[H_3O^+]} = 10^{-2,40}

[H₃O⁺] = <em>0,004M</em>

Thus:

[X⁻] = 0,004M

And:

[HX] = 0,089M - 0,004M = <em>0,085M</em>

ka = \frac{[0,004][0,004]}{[0,085]}

ka = 1,88x10⁻⁴

And <em>pka = 3,73</em>

b. As pka + pkb = 14,00

pkb = 14,00 - 3,73

<em>pkb = 10,27</em>

I hope it helps!

4 0
2 years ago
Identify the compounds below as a brønsted-lowry acid or lewis acid by clicking and dragging it into the correct column
Vesna [10]
There are many types of acid or bases. Based on the Bronsted-Lowry definition,

* A Bronsted-Lowry acid is a proton donor
* A Bronsted-Lowry base is a proton acceptor

Take this reaction for example:
HCl(aq)+ N<span>H</span>₃(aq)→N<span>H</span>⁴⁺(aq)+C<span>l</span>⁻(aq<span>)
</span>
HCl donates a proton, so it is a Bronsted-Lowry acid. Consequently, ammonia accepts this proton, so it is the Bronsted-Lowry base.
3 0
2 years ago
Read 2 more answers
(f) what is the observed rotation of 100 ml of a solution that contains 0.01 mole of d and 0.005 mole of l? (assume a 1-dm path
never [62]
<span>Answer: .01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L. so .01/.015 to .005/.015 ~ 67% D to 33% L. And thus, the enantiomer excess will be 34%.</span>
4 0
2 years ago
Read 2 more answers
A graduated cylinder is filled to 10.0 mL with water and a piece of granite is placed in the cylinder displacing the level to 23
Rus_ich [418]

Answer:

13.7 cubic centimeters

Explanation:

23.7-10=13.7mL

8 0
2 years ago
A standard backpack is approximately 30cm x 30cm x 40cm. Suppose you find a hoard of pure gold while treasure hunting in the wil
Blizzard [7]

Explanation:

The dimensions of a standard backpack is 30cm x 30cm x 40cm

The mass of an average student is 70 kg

We know that, the density of gold is 19.3 g/cm³.

Let m be the mass of the backpack. So,

\text{density}=\dfrac{\text{mass}}{\text{volume}}\\\\m=d\times V\\\\m=19.3\ g/cm^3\times (30\times 30\times 40)\ cm^3\\\\m=694800\ g\\\\\text{or}\\\\m=694.8\ kg\approx 700\ kg

An average student has a mass of 70 kg. If we compare the mass of student and mass of backpack, we find that the backpack is 10 times of the mass of the student.

8 0
2 years ago
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