What is the charge on an electron? -1 0 +1 -2
2 answers:
_____________________________
As we know that the charge of the electron is equivalent to the magnitude of the elementary charge that is (e) but bearing a negative sign.
Since, the value of the elementary charge is roughly
then, the charge of the electron is :
Here, C is the unit of charge is Coulomb.
So, the charge on an electron is (-1).
_____________________________
As we know that the charge of the electron is equivalent to the magnitude of the elementary charge that is (e) but bearing a negative sign.
Since, the value of the elementary charge is roughly
then, the charge of the electron is :
Here, C is the unit of charge is Coulomb.
So, the charge on an electron is (-1).
_____________________________
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Answer:
The weight percent in the sample is 17,16%
Explanation:
The dissolution of the Ce(IV) salt provides free Ce⁴⁺ that reacts, thus:
2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻
I₃⁻ + 2S₂O₃²⁻ → 3I⁻ + S₄O₆²⁻
The moles in the end point of S₂O₃⁻ are:
0,01302L×0,03247M Na₂S₂O₃ = 4,228x10⁻⁴ moles of S₂O₃²⁻.
As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:
4,228x10⁻⁴ moles of S₂O₃⁻× = <em>2,114x10⁻⁴ moles of I₃⁻</em>
As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:
2,114x10⁻⁴ moles of I₃⁻× = <em>4,228x10⁻⁴ moles of Ce(IV)</em>.
These moles are:
4,228x10⁻⁴ moles of Ce(IV)× = 0,05924 g of Ce(IV)
As was taken an aliquot of 25,00mL from the solution of 250,0mL:
0,05924 g of Ce(IV)× =0,5924g of Ce(IV) in the sample
As the sample has 3,452g, the weight percent is:
0,5924g of Ce(IV) / 3,452g × 100 = <em>17,16 wt%</em>
I hope it helps!
I’m not an expert and I haven’t taken chemistry in like a year but i believe you use stoichiometry to do this equation. It should look something like this.
