<u>Answer:</u> The concentration of
in the solution is 
<u>Explanation:</u>
The given cell is:

Half reactions for the given cell follows:
<u>Oxidation half reaction:</u>
( × 3)
<u>Reduction half reaction:</u>
( × 2)
<u>Net reaction:</u> 
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the
of the reaction, we use the equation:

Putting values in above equation, we get:

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:
![E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E6%7D%7B%5BAu%5E%7B3%2B%7D%5D%5E2%7D)
where,
= electrode potential of the cell = 1.23 V
= standard electrode potential of the cell = +1.50 V
n = number of electrons exchanged = 6
![[Au^{3+}]=?M](https://tex.z-dn.net/?f=%5BAu%5E%7B3%2B%7D%5D%3D%3FM)
![[H^{+}]=1.0M](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%3D1.0M)
Putting values in above equation, we get:
![1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})](https://tex.z-dn.net/?f=1.23%3D1.50-%5Cfrac%7B0.059%7D%7B6%7D%5Ctimes%20%5Clog%28%5Cfrac%7B%281.0%29%5E6%7D%7B%5BAu%5E%7B3%2B%7D%5D%5E2%7D%29)
![[Au^{3+}]=1.87\times 10^{-14}M](https://tex.z-dn.net/?f=%5BAu%5E%7B3%2B%7D%5D%3D1.87%5Ctimes%2010%5E%7B-14%7DM)
Hence, the concentration of
in the solution is 
Answer:
A) 14. 25 × 10²³ Carbon atoms
B) 34.72 grams
Explanation:
1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.
The sample has 3.84 × 10²⁴ H atoms.
If 8 atoms of Hydrogrn are present in 1 molecule of propane.
3.84 × 10²⁴ H atoms are present in

<u>= 4.75 × 10²³ molecules of Propane</u>.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
No. of Carbon atoms in 1 molecule of propane = 3
=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³
<u>= 14.25 × 10²³ </u>
<u>________________________________________</u>
<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>
= 3 × 12 + 8 × 1
= 36 + 8
= 44 g
1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.
=> 6.02 × 10²³ molecules of Propane weigh = 44 g
=> 4. 75 × 10²³ molecules of Propane weigh =



<u>= 34.72 g</u>
Answer : The number of iron atoms present in each red blood cell are, 
Explanation :
First we have to calculate the moles of iron.

Now we have to calculate the number of iron atoms.
As, 1 mole of iron contains
number of iron atoms
So, 0.0519 mole of iron contains
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 = 
Number of iron atoms are present in each red blood cell = 
Number of iron atoms are present in each red blood cell = 
Therefore, the number of iron atoms present in each red blood cell are, 
An atom is the smallest particle of an element that can take part in a chemical reaction.
An atom is made up of energy levels that contain electrons which are negatively charged and the nucleus which contains neutrons and protons that are negatively charge .
Due the positive charge of the nucleus of an atom, an atom always want to attract its electrons and keep them near it however it weakly attracts the other electrons of a nearby atom.