Answer:
Explanation:
2Al(s) + 3 I₂(s) → 2 Al⁺³ + 6 I⁻
Aluminium is oxidised and iodine is reduced .
so cell potential = Ereduction - Eoxidation
Al⁺³ + 3e = Al - 1.66 V
I₂ + 2 e = 2 I⁻ 0.54 V
= .54 - ( - 1.66 )
= 1.66 + .54
= 2.2 V
Answer:
The temperature at which a substance changes from the liq- uid state to the solid state is called the freezing point. The freez- ing point of the liquid state of a substance is the same temperature as the melting point of the solid state. For example, solid water melts at 0°C and liquid water freezes at 0°C.
Explanation:
"LEO the lion goes GER" means based upon the oxidation and reduction reactions.
<h3>What is oxidation and reduction reaction?</h3>
Oxidation reactions are those reactions in which lossing of electrons are taking place from the substrate molecule and in reduction process gaining of electrons is done by substrate molecules.
The phrase "LEO the lion goes GER" is used in the context of oxidation and reduction reactions. Meaning of the terms:
LEO → Lose electrons = oxidation
GER → Gain electrons = reduction
Hence the given line is based on oxidation and reduction.
To know more about oxidation and reduction, visit the below link:
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Answer:
17.65 grams of O2 are needed for a complete reaction.
Explanation:
You know the reaction:
4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O
First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values of the atomic mass of each element that form the compounds:
- N: 14 g/mol
- H: 1 g/mol
- O: 16 g/mol
So, the molar mass of the compounds in the reaction is:
- NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
- O₂: 2*16 g/mol= 32 g/mol
- NO: 14 g/mol + 16 g/mol= 30 g/mol
- H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol
By stoichiometry, they react and occur in moles:
- NH₃: 4 moles
- O₂: 5 moles
- NO: 4 moles
- H₂O: 6 moles
Then in mass, by stoichiomatry they react and occur:
- NH₃: 4 moles*17 g/mol= 68 g
- O₂: 5 moles*32 g/mol= 160 g
- NO: 4 moles*30 g/mol= 120 g
- H₂O: 6 moles*18 g/mol= 108 g
Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

mass of O₂≅17.65 g
<u><em>17.65 grams of O2 are needed for a complete reaction.</em></u>