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:
d. ii.
Explanation:
Hello!
In this case, according to the displayed decimal placed by each balance, we can notice that:
i. Displays 1 decimal place.
ii. Displays 2 decimal places.
iii. Displays 4 decimal places.
iv. Displays 3 decimal places.
Thus, since the measured-out masses have two decimal places each, we infer that the used balance was d. ii as it displays two decimal places.
Best regards!
Abca I think it’s right not sure
