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kotegsom [21]
3 years ago
9

All of the cells in the body need oxygen. Hemoglobin molecules in red blood cells transport oxygen through the bloodstream. Oxyg

en is loaded onto hemoglobin molecules in the lungs and unloaded from the hemoglobin molecules in the tissues. What drives the unloading of oxygen from hemoglobin molecules in the tissues? the high partial pressure of carbon dioxide in the tissues the low partial pressure of carbon dioxide in the tissues the low partial pressure of oxygen in the tissues the high partial pressure of oxygen in the tissues
Chemistry
1 answer:
tankabanditka [31]3 years ago
6 0

Answer:

High partial pressure of oxygen in the tissues.

Explanation:

Hemoglobin is a protein found in the red blood cells that is responsible for the transportation of oxygen from the lungs to the rest of the tissues in the body. The partial pressure determines whether oxygen is loaded or unloaded into the hemoglobin.

When there is lack of oxygen in our bodies, we experience hypoxia.

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Two liters of oxygen (O2) is collected over water at 20oC. If the total pressure in the container is 95.00 kPa, what is the part
garri49 [273]

Answer:

THE PARTIAL PRESSURE OF OXYGEN GAS IN THE CONTAINER IS 92.67kPa WHICH IS OPTION B.

Explanation:

To calculate the partial pressure of oxygen gas collected over water, we use

Ptotal = Poxygen + P water

It is worthy to note that when oxygen is collected over water, it is mixed with water vapor and the total pressure in the container will be the sum of the pressure exerted by the oxygen gas and that of the water vapor at that given temperature.

At 20 C, the vapor pressure of water as given in the question is 2.33 kPa.

Using the above formula,

Ptotal = Poxygen + P water

Substituting for Poxygen, we have;

Poxygen = Ptotal - P water vapor

P oxygen = 95 .00 kPa - 2.33 kPa

P oxygen = 92.67 kPa.

The partial pressure of oxygen gas in the container is hence, 92.67kPa.

3 0
3 years ago
40 points plz help quick!! Order the interactions from strongest to weakest.
aleksley [76]

Answer:

2,5,4,1,3

Explanation:

7 0
3 years ago
Read 2 more answers
A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric ac
UkoKoshka [18]

Answer:

Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

Explanation:

The pH of the buffer solution before the addition of HCl is:

pH = pKa + log(\frac{[KF]}{[HF]})

pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04  

The hydrochloric acid added will react with the potassium fluoride as follows:

H₃O⁺(aq)  +  F⁻(aq) ⇄   HF(aq) + H₂O(l)

The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:

\eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol

\eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles

The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:

\eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles  

Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.  

We can calculate the pH after the addition of HCl:

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq)    (1)

The number of moles of HF after the reaction of KF with HCl is:

\eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles

And the concentration of HF after the reaction of KF with HCl is is:

C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L

Now, from the equilibrium of equation (1) we have:

Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]}

Ka = \frac{x^{2}}{0.531 - x}  (2)

By solving equation (2) for x we have:

x = 0.0187

Finally, the pH after the addition of HCl is:

pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73

Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

I hope it helps you!

8 0
4 years ago
A photon of light possesses 5 x 10^-19 J of energy. Calculate its frequency
saveliy_v [14]

Answer:

The frequency of photon is 0.75×10¹⁵ s⁻¹.

Explanation:

Given data:

Energy of photon = 5×10⁻¹⁹ J

Frequency of photon = ?

Solution:

Formula;

E = hf

h = planck's constant = 6.63×10⁻³⁴ Js

5×10⁻¹⁹ J =  6.63×10⁻³⁴ Js ×f

f =  5×10⁻¹⁹ J / 6.63×10⁻³⁴ Js

f = 0.75×10¹⁵ s⁻¹

The frequency of photon is 0.75×10¹⁵ s⁻¹.

4 0
3 years ago
The combustion of methane(CH4) produces carbon dioxide and water. Assume that 2.0 mol of CH4 burned in the presence of excess ai
harkovskaia [24]

Answer:

The answer to your question is 98.9 %

Explanation:

Data

moles of methane = CH₄ = 2.0

excess air

Percent yield = ?

mass of CO₂ = 87 g

- Balanced chemical reaction

                CH₄  +  2O₂   ⇒  CO₂  +  2H₂O

      Reactants     Elements       Products

             1                    C                   1

             4                   H                   2

             4                   O                   2

- Calculate the molar mass of CH₄

CH₄ = 12 + 4 = 16 g

- Convert the moles to mass

                     16 g of CH₄ ----------------- 1 mol

                       x                -----------------  2 moles

                       x = (2 x 16) / 1

                       x = 32 g of CH₄

-Calculate the theoretical formation of CO₂

                     16 g of CH₄ ----------------- 44 g of CO₂

                     32 g of CH₄ ----------------  x

                            x = (32 x 44) / 16

                            x = 88 g of CO₂

-Calculate the Percent yield

     Percent yield = Actual yield/Theoretical yield x 100

     Percent yield = 87/88 x 100

    Percent yield = 98.9 %

3 0
3 years ago
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