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masha68 [24]
3 years ago
7

Carbon dioxide affects O2binding to Hb by: a) CO2forming iron carbonate with the heme-iron atom. b) Hb competing with carbonic a

nhydride for CO2. c) directly binding to heme-Fe in the oxygen binding site of hemoglobin. d) forming HCO3- that combines with H+ to increase CO2binding. e) forming H+ HCO3- where the H is an antagonist to oxygen binding to Hb.
Chemistry
1 answer:
exis [7]3 years ago
5 0

Answer:

The correct answer is c) directly binding to heme Fe in the oxygen binding site of the hemoglobin.

Explanation:

Carbon dioxide is the waste product of cellular respiration.Carbon dioxide bind to hemoglobin to form carbaminohemoglobin in which the hemoglobin exist as T state or Tensed state.

        Carbon dioxide interact with the oxygen binding site of hemoglobin thus decreasing the affinity of hemoglobin for oxygen.

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Answer:

When carbon is burned in air, it forms carbon dioxide gas and releases a large amount of heat and some light:

C+O2= CO2+ heat+ light

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For the reaction: Cr2O7^2- + NH4^+ → Cr2O3 + N2, which is the oxidizing agent?
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(NH4)2Cr2O7

Explanation:

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By which process is sediment laid down?<br> its a test it's due at 10!
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A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is
Charra [1.4K]

Answer:

[H^{+}] = 0.761 \frac{mol}{L}

[OH^{-}]=1.33X10^{-14}\frac{mol}{L}

pH = 0.119

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L)  \\ n_{H^{+} } from HNO_{3}  = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

n_{H^{+} } from HCl = (5.00)(0.093)

n_{H^{+} } from HCl = 0.465 mol

n_{H^{+} } from HNO_{3}  = (8.00)(0.037)

n_{H^{+} } from HNO_{3}  = 0.296 mol

n_{H^{+}(total) } = 0.296 + 0.465

n_{H^{+}(total) } = 0.761 mol

For molar concentration of hydrogen ions:

[H^{+}]  = \frac{n_{H^{+}}(mol)}{V(L)}

[H^{+}] = \frac{0.761}{1.00}

[H^{+}] = 0.761 \frac{mol}{L}

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

K_{w} = [H^{+} ][OH^{-} ]

[OH^{-}]=\frac{Kw}{[H^{+}] }

[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }

[OH^{-}]=1.33X10^{-14}\frac{mol}{L}

The pH of the solution can be measured by the following formula:

pH = -log[H^{+} ]

pH = -log(0.761)

pH = 0.119

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