You have to use everything that is given since you have to know which is the limiting reactant. We find the limiting reactant by calculating the number of moles of each reactant and compare the number of moles. The limiting reactant would be the one that is consumed fully by the reaction.
<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.
<u>Explanation:</u>
We are given:
The substance having highest positive 
potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.
Aluminium will undergo oxidation reaction and will get oxidized.
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:
Hence, the standard electrode potential of the cell is 4.53 V.
<span><span>There is no formula. The speed of light is a fundamental constant which appears in other formulas but there’s no formula to compute the numerical value.Well, actually, that’s not quite right. The numerical value in meters per second is known exactly, because we use the speed of light to define the meter. It is: <span><span><span>c=299,792,458 m/s</span><span>c=299,792,458 m/s</span></span>
</span>. Exactly. But the thing is — this value is purely an artifact of our unit system. Other unit systems will give other values, so the number value is entirely arbitrary.</span></span>
If you are provided with Cation and an Anion with different oxidation states, then there ratio in the formula unit is adjusted as such that the oxidation number of one ion is set the coefficient of other ion and vice versa,
Example:
Let suppose you are provided with A⁺² and B⁻¹, so multiply A by 1 and B by 2 as follow,
A(B)₂
In statement we are given with Co⁺³ and SO₄⁻², so multiply Co⁺³ by 2 and SO₄⁻² by 3, hence,
Co₂(SO₄)₃
Result:
Co₂(SO₄)₃ is the correct answer.
A<span>. </span>
<span>the change in carbon dioxide levels in the atmosphere
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hope that's right and hope i helped
