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Ira Lisetskai [31]
3 years ago
10

Iron is biologically important in the transport of oxygen by red blood cells from the lungs to the various organs of the body. I

n the blood of an adult human, there are approximately 2.64 × 1013 red blood cells with a total of 2.90 g of iron. On the average, how many iron atoms are present in each red blood cell? (molar mass Fe = 55.85 g/mol)
Chemistry
1 answer:
Aneli [31]3 years ago
6 0

Answer : The number of iron atoms present in each red blood cell are, 1.077\times 10^9

Explanation :

First we have to calculate the moles of iron.

\text{Moles of iron}=\frac{\text{Mass of iron}}{\text{Molar mass of iron}}=\frac{2.90g}{55.85g/mole}=0.0519moles

Now we have to calculate the number of iron atoms.

As, 1 mole of iron contains 6.022\times 10^{23} number of iron atoms

So, 0.0519 mole of iron contains 0.0519\times 6.022\times 10^{23}=3.125\times 10^{22} number of iron atoms

Now we have to calculate the number of iron atoms are present in each red blood cell.

Number of iron atoms are present in each red blood cell = \frac{\text{Number of iron atoms}}{\text{Total number of red blood cells}}

Number of iron atoms are present in each red blood cell = \frac{3.125\times 10^{22}}{2.90\times 10^{13}}

Number of iron atoms are present in each red blood cell = 1.077\times 10^9

Therefore, the number of iron atoms present in each red blood cell are, 1.077\times 10^9

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Molecule of water contains hydrogen and oxygen in a 1:8 ratio by mass. This is a statement of __________.A) the law of multiple
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Answer:

The correct answer is B.

Explanation:

The molecule of water has 2 atoms of hydrogen and 1 atom of oxygen.

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This illustrates the law of definite proportions which is also known as law of constant compositions .

The law states that 'the elements combining to form compound always combine in a fixed ratio by their mass.'

Whereas :

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What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
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Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

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d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

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Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

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