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sveticcg [70]
3 years ago
10

Which of the following is used up during cellular respiration?

Biology
2 answers:
svet-max [94.6K]3 years ago
7 0

The correct answer is C. Oxygen

Hope this helps

-AaronWiseIsBae

ra1l [238]3 years ago
7 0

The correct answer is C

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Art [367]

Answer: V(t)=5000(9/10)^t

Explanation:

If the car loses 1/10 of its value each year, that means 9/10 of the value remains each year.

So each year, the car's value is multiplied by a factor of 9/10 or 0.9

If we start with the initial value, $5000 and keep multiplying by 9/10 his function gives us the car's value t years from now:

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3 years ago
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Answer:

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2 years ago
What organelle do frog RBCs have that human RBCs do not and explain why?
Lina20 [59]
Frog RBCs contain a DNA-bearing nucleus that is visible in the center of the cell. Human RBCs do not possess nucleus along with other cell organelles such as mitochondria, Golgi apparatus and endoplasmic reticulum in order to accommodate greater amount of haemoglobin in the cells. Denucleation of rbcs is an adaptation, Which makes the mammalian red blood cell effective at transporting oxygen/eliminating CO2.
4 0
3 years ago
Advertising objectives can be classified based on three primary purposes. what are the three​ purposes?
Viefleur [7K]
The three primary purposes that advertising objectives can be classified by area; To inform; this is in the case of a new product where the company needs to inform the customers and create awareness on the existence of a new product. Secondly; To persuade; once you are in a competitive environment there is need to persuade the customers to buy your product. Lastly, To remind; this is especially important for a mature brand. 
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3 years ago
The concentration of N2 in blood at 37 °C (body temperature) and atmospheric pressure (partial pressure of N2 = 0.80 atm) is 0.0
devlian [24]

Answer:

volume  = 0.285 L

Explanation:

Henry constant is given as

Henry constant  =  \frac{C}{P}

Where C  is concentration

P is atmospheric pressure

Henry\ constant = \frac{0.00056}{0.80} = 0.0007 M/atm

when atmospheric pressure 4 atm

solubility = K_H\times P

              = 0.0007 \times 4 = 0.0028 M

In 5 litere blood , moles of N_2 = Molarity \times volume

                                                     = 0.0028 \times 5 = 0.014

At surface moles  = 0.0056 \times 5 = 0.0028

Moles of N_2 release = 0.014 - 0.0028 = 0.0112

Mass of N_2 = 0.0112 \times 28 = 0.314 g

T = 37 + 273 = 310 K

VOLUME = \frac{nRT}{P}

= \frac{0.0112\times 0.0821\times 310}{1}

volume  = 0.285 L

6 0
3 years ago
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