Answer:
1 billion molecules O₂
Explanation:
From my research, a human red blood cell contains approximately 270 million hemoglobin molecules.
A hemoglobin molecule contains four heme groups, <em>each of which has an iron ion forming a coordination complex that carries every dioxygen molecule. </em>Therefore for each hemoglobin molecule, we will have 4 dioxygen molecules. The heme groups are responsible for the transport of every dioxygen and other diatomic gases.
Hence, the number of O₂ molecules in a red blood cell saturated with 100% will be:

So, the correct answer is 1 billion of O₂ molecules.
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Answer:
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Answer: D is right
Explanation: One mole contains 6.0225 ·10^23 molecules
Despite of the substance
<u>Answer:</u> The phase change process in which solids gets converted to gases is sublimation.
<u>Explanation:</u>
For the given options:
<u>Option a:</u> Condensation
It is a type of process in which phase change occurs from gaseous state to liquid state at constant temperature.

<u>Option b:</u> Melting
It is a type of process in which phase change occurs from solid state to liquid state at constant temperature.

<u>Option c:</u> Sublimation
It is a type of process in which phase change occurs from solid state to gaseous state without passing through the liquid state at constant temperature.

<u>Option d:</u> Deposition
It is a type of process in which phase change occurs from gaseous state to solid state without passing through the liquid state at constant temperature.

Hence, the phase change process in which solids gets converted to gases is sublimation.
Answer:
Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.
Explanation:
The unit of k is s⁻¹.
The order of the reaction = first order.
First order reaction: A first order reaction is a reaction in which the rate of reaction depends only the value of the concentration of the reactant.
![-\frac{d[A]}{dt} =kt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3Dkt)
[A] = the concentration of the reactant at time t
k= rate constant
t= time
Here k= 4.70×10⁻³ s⁻¹
t= 4.00
[A₀] = initial concentration of reactant = 0.700 M
![-\frac{d[A]}{dt} =kt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3Dkt)
![\Rightarrow -\frac{d[A]}{[A]}=kdt](https://tex.z-dn.net/?f=%5CRightarrow%20-%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%3Dkdt)
Integrating both sides
![\Rightarrow\int -\frac{d[A]}{[A]}=\int kdt](https://tex.z-dn.net/?f=%5CRightarrow%5Cint%20-%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%3D%5Cint%20kdt)
⇒ -ln[A] = kt +c
When t=0 , [A] =[A₀]
-ln[A₀] = k.0 + c
⇒c= -ln[A₀]
Therefore
-ln[A] = kt - ln[A₀]
Putting the value of k, [A₀] and t
- ln[A] =4.70×10⁻³×4 -ln (0.70)
⇒-ln[A]= 0.375
⇒[A] = 0.686
Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.