What is the source of most of the salt in the oceans?

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

Aneli [31]

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, $1.077\times 10^9$

6 0

2 years ago

Which example provides a complete scientific description of an object in motion? Question 3 options: The marble changed position

Contact [7]

The answer is the Car Traveling North... According to me

3 0

3 years ago

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1.72 moles of NOCI were placed in a 2.50 L reaction chamber

elena-14-01-66 [18.8K]

The equilibrium constant, Kc=0.026

<h3>Further explanation</h3>

Given

1.72 moles of NOCI

1.16 moles of NOCI remained

2.50 L reaction chamber

Reaction

2NOCI(g) = 2NO(g) + Cl2(g).

Required

the equilibrium constant, Kc

Solution

ICE method

2NOCI(g) = 2NO(g) + Cl2(g).

I 1.72

C 0.56 0.56 0.28

E 1.16 0.56 0.28

Molarity at equilibrium :

NOCl :

$\tt \dfrac{1.16}{2.5}=0.464$

NO :

$\tt \dfrac{0.56}{2.5}=0.224$

Cl2 :

$\tt \dfrac{0.28}{2.5}=0.112$

$\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026$

4 0

3 years ago

In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n

artcher [175]

Stoichiomety:

1 moles of C + 1 mol of O2 = 1 mol of CO2

multiply each # of moles times the atomic molar mass of the compund to find the relation is weights

Atomic or molar weights:

C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol

Stoichiometry:

12 g of C react with 32 g of O2 to produce 44 g of CO2

Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen

And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.

You cannot obtain 72 g of CO2 from 18 g of C.

May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.

3 0

3 years ago

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What is the molar mass of CaCl2?

alekssr [168]

Molar mass of CaCl2 = 40+ ( 35.5 ×2)=110

Mr of Ca(OH)2 = 40+ (16+1)×2 =74

%of Ca = (40÷ 74)×10=...

1 m = 100cm...
1cm = (1÷100) m
So 45.5 cm = 45.5 ×(1÷100) =....

1km = 1000m
1m = 100 cm
1cm =10mm
So 1km = 1000×100×10 mm
Now convert

7 0

2 years ago