First, we write the half equations for the reduction of the chemical species present:
Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V
In order to determine the potential of the cell, we find the difference between the two values. For this:
E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V
The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
Given:3.40g sample of the steel used to produce 250.0 mLSolution containing Cr2O72−
Assuming all the Cr is contained in the BaCrO4 at the end.
(0.145 g BaCrO4) / (253.3216 g BaCrO4/mol) x (250.0 mL / 10.0 mL) x (1 mol Cr / 1 mol BaCrO4) x (51.99616 g Cr/mol / (3.40 g) = 0.219 = 21.9% Cr
Answer:
Metals have one or two electrons in their outermost shell
C. 1-2
Explanation:
- Metals have low ionisation energy because they easily looses the outermost electrons
- They have only one- two electrons in the outer most shell.
- They loose these electron to form charged species called cation.
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