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Anon25 [30]
3 years ago
15

If a solution contains 1 x 106 hb molecules, how many o2 binding sites are available? if the solution of hb is 50% saturated wit

h o2, how many binding sites are occupied by o2? if the hb solution is 23% saturated with o2, how many binding sites are occupied by o2?
Chemistry
1 answer:
Liula [17]3 years ago
8 0

Actually each molecule of Hb can bind four oxygen molecules. So given 1 x 10^6 Hb, so there are 4 x 10^6 binding sites available.

If only 50% of the Hb are bound, so this means that 2 x 10^6 binding sites are occupied by O2. However we must remember that hemoglobin binding is an all-or-nothing phenomenon. Meaning that each molecule of Hb binds four molecules of oxygen or zero at all.

<span>If 23% is saturated, so the number of binding sites occupied is 0.92 x 10^6 binding sites or 9.2 x 10^5 binding sites.</span>

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When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were
uysha [10]

Answer:

\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}

(b) Mass of H

\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}

(c)Mass of Fe

(i)In 0.4131g of X

\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}

(ii) In 1.2383 g of X

\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}

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Explanation:

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