Answer:
(a) X electrode
(b) Y electrode
(c) Y electrode
(d) X electrode
(e) Y electrode
Explanation:
<em>A galvanic (voltaic) cell has the generic metals X and Y as electrodes. X is more reactive than Y, that is, X more readily reacts to form a cation than Y does.</em>
In the X electrode occurs the oxidation whereas in the Y electrode occurs the reduction.
Oxidation: X(s) → X⁺ⁿ(aq) + n e⁻
Reduction: Y⁺ˣ(aq) + x e⁻ → Y(s)
<em>Classify the descriptions by whether they apply to the X or Y electrode.
</em>
<em>(a) anode.</em> Is where the oxidation takes place (X electrode).
<em>(b) cathode.</em> Is where the reduction takes place (Y electrode).
<em>(c) electrons in the wire flow toward.</em> Electrons in the wire flow toward the cathode (Y electrode).
<em>(d) electrons in the wire flow away.</em> Electrons in the wire flow away from the anode (X electrode).
<em>(e) cations from salt bridge flow toward.</em> Cations from the salt bridge flow toward the cathode (Y electrode) to maintain the electroneutrality.
Answer:
pH = 11.60
Explanation:
When we add a base, we are increasing [OH⁻], so the pH will be totally basic.
pH of water = 7
Basic pH > 7
We are adding 2 M . 1 mL = 2 mmoles
2 mmoles of OH⁻ are contained in 501 mL of total volume.
[OH⁻] = 2 mmol / 501 mL = 3.99×10⁻³ M
- log 3.99×10⁻³ M = 2.39 → pOH
pH = 14 - pOH → 11.61
Water equilibrium
2H₂O ⇄ H₃O⁺ + OH⁻ Kw = 1×10⁻¹⁴
NaOH → Na⁺ + OH⁻
The answer is (1) 1.3 M. The first thing you need to do is to convert the unit of gram to mole. The mol number of LiF is 52/26=2 mol. Then using the volume to calculate the molarity: molarity=2/1.5=1.3 M.
Answer:
If you dissolve 58.44g of NaCl in a final volume of 1 liter, you have made a 1M NaCl solution, a 1 molar solution.
Explanation:
