Respuesta :
Answer:
10.6 mA
Explanation:
t = time interval = 1.00 s
q = magnitude of charge on each ion = 1.6 x 10⁻¹⁹ C
n₁ = number of Na⁺ ions = 2.68 x 10¹⁶
q₁ = charge due to Na⁺ ions = n₁ q = (2.68 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.004288 C
n₂ = number of Cl⁻ ions = 3.92 x 10¹⁶
q₂ = charge due to Cl⁻ ions = n₂ q = (3.92 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.006272 C
i₁ = Current due to Na⁺ ions = [tex]\frac{q_{1}}{t}[/tex] = [tex]\frac{0.004288}{1}[/tex] = 0.004288 A
i₂ = Current due to Cl⁻ ions = [tex]\frac{q_{2}}{t}[/tex] = [tex]\frac{0.006272}{1}[/tex] = 0.006272 A
Current passing between the electrodes is given as
i = i₁ + i₂
i = 0.004288 + 0.006272
i = 0.01056 A
i = 10.6 x 10⁻³ A
i = 10.6 mA
A current of 10.56 mA passes through a sodium chloride solution causing 2.68 × 10¹⁶ Na⁺ ions and 3.92 × 10¹⁶ Cl⁻ ions to arrive at their respective electrodes in 1.00 s.
What is an electric current?
An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space.
- Step 1: Calculate the electric current due to Na⁺ ions
2.68 × 10¹⁶ Na⁺ ions (1.60 × 10⁻¹⁹ C/ion) arrive at the negative electrode in 1.00 s.
I₁ = 2.68 × 10¹⁶ ion × (1.60 × 10⁻¹⁹ C/ion)/ 1.00 s × (10³ mA/1 A) = 4.29 mA
- Step 2: Calculate the electric current due to Cl⁻ ions
3.92 × 10¹⁶ Cl⁻ ions (1.60 × 10⁻¹⁹ C/ion) arrive at the positive electrode in 1.00 s.
I₂ = 3.92 × 10¹⁶ ion × (1.60 × 10⁻¹⁹ C/ion)/ 1.00 s × (10³ mA/1 A) = 6.27 mA
- Step 3: Calculate the total current passing between the electrodes.
I = I₁ + I₂ = 4.29 mA + 6.27 mA = 10.56 mA
A current of 10.56 mA passes through a sodium chloride solution causing 2.68 × 10¹⁶ Na⁺ ions and 3.92 × 10¹⁶ Cl⁻ ions to arrive at their respective electrodes in 1.00 s.
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