Answer :
The Nernst equation :
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Anode]}{[Cathode]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BAnode%5D%7D%7B%5BCathode%5D%7D)
where,
= standard cell potential
n = number of electrons in oxidation-reduction reaction
F = Faraday constant = 96500 C
R= gas constant = 8.314 J/Kmol
T = temperature
[Anode] = anodic ion concentration
[Cathode] = cathodic ion concentration
The number of moles that are contained in the given mass of propane (
is 1.7143 moles.
<u>Given the following data:</u>
- Mass of propane = 75.6 grams.
<u>Scientific data:</u>
- The molar mass of propane = 44.1 g/mol.
To calculate the number of moles that are contained in the given mass of propane (
):
<h3>How to calculate the moles of a compound.</h3>
In this exercise, you're required to determine the number of moles of propane that are contained in the given sample:
Mathematically, the number of moles contained in a chemical compound is given by this formula:

Substituting the given parameters into the formula, we have;

Number of moles = 1.7143 moles.
Read more on number of moles here: brainly.com/question/3173452
Explanation:
The balanced chemical equation of the reaction is:

From the balanced chemical equation,
1 mole of propane forms ------ 3 mol. of
gas.
The molar mass of propane is 44.1 g/mol.
One mole of any gas at STP occupies --- 22.4 L.
Hence, 44 g of propane forms (3x22.4 L=) 67.2 L of CO2 gas at STP.
Answer:
Thus, 67.2 L of CO2 is formed at STP.
Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...
Using an ionic chloride and Phosphoric acid
H3PO4 + NaCl ==> HCl + NaH2PO4
H3PO4 + NaI ==> HI + NaH2PO4
H2SO4 + NaCl ==> HCl + NaHSO4
This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.
The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.
2I-² === I2 + 2e-