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worty [1.4K]
3 years ago
11

Which three of the following statements accurately describe the blood buffering system in humans? the blood buffering system . .

.
Biology
1 answer:
Bogdan [553]3 years ago
8 0
The truth about blood buffering is that 1). mantains the ph of blood near to 7.4. 2) utilizes the H2CO3/HCO3– conjugate acid/base pair and 3) is facilitated by the enzyme carbonic anhydrase, which interconverts carbon dioxide and water to carbonic acid. Have in mind that the buffer is written as the following: <span>CO2(aq) + H2O(l) <==> H+(aq) + HCO3^-(aq) </span>
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An Asian elephant has a weight of 52,900 newtons. This enormous weight is balanced on four feet. In a circus act, these elephant
Aleonysh [2.5K]

Answer:

Option B, 18.11 \frac{N}{cm^{2}}

Explanation:

Given-

The average surface area of one foot pad of an elephant is 2920 cm^{2}\\

Weight of an Asian elephant is 52900 newton

As we know, pressure is equal to total weight divided by total area of the surface on which it is concentrated.

Thus, pressure on one foot  pad of elephant is \frac{52900}{2920} \frac{N}{cm^{2}} \\

= 18.11\frac{N}{cm^{2}}

Hence, option B is correct answer

7 0
2 years ago
The medical term ____________________ describes a pus-filled lesion on the eyelid resulting from an infection in a sebaceous gla
zhuklara [117]

Answer:

the medical term is hordeolum

8 0
2 years ago
Epiphytes that live on trees in a tropical rain forest show this type of symbiosis?
vovangra [49]
Epiphytes show commensalism.
8 0
3 years ago
A man is brachydactylous (very short fingers; rare autosomal dominant), and his wife is not. Both can taste the chemical phenylt
netineya [11]

Answer:

The correct answer is -

A) BbTt x bbTt

B) .5^4 or 1/16 or 0.0625

Explanation:

As it is given that Brachydactylus and PTC tasters both traits are autosomal dominant conditions which mean only one allele would be enough.

For Branchydactylus and For tasting :  man and women will be heterogeneous.

Hence,

The Genotype of man = BbTt

The Genotype of wife = bbTt

b. Answer = 0.0625

For Branchydactylus:

Bb X bb

Possible genotypes:

B b

b Bb( Brachydactylus) bb(normal)

b Bb (Brachydactylus) bb (normal)

The probability of a single child being Branchydactylus = 2/4 = 0.5

So,  Probability of all 4 child being Branchydactylus = .5 x .5 x .5 x .5 = 0.0625

3 0
2 years ago
¿que función desempeña el hospedador en el desarrollo de una enfermedad infectocontagiosa? ¿Que otros insectos, además del mosqu
olga2289 [7]

Answer:

origami gamo

Explanation:

8 0
2 years ago
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