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arsen [322]
3 years ago
15

A galvanic (voltaic) cell has the generic metals X and Y as electrodes. X is more reactive than Y , that is, X more readily reac

ts to form a cation than Y does. Classify the descriptions by whether they apply to the X or Y electrode.
Chemistry
1 answer:
Margaret [11]3 years ago
7 0

Answer:

(1) cathode: Y

(2) anode X

(3) electrons in the wire flow toward: Y

(4) electrons in the wire flow away from: X

(5) anions from the salt bridge flow toward X

(6) cations from the salt bridge flow toward Y

(7) gains mass: Y

(8) looses mass X

Explanation:

The voltaic cell uses two different metal electrodes, each in an electrolyte solution. The anode will undergo oxidation and the cathode will undergo reduction. The metal of the anode will oxidize, going from an oxidation state of 0 (in the solid form) to a positive oxidation state, and it will become an ion. At the cathode, the metal ion in the solution will accept one or more electrons from the cathode, and the ion’s oxidation state will reduce to 0. This forms a solid metal that deposits on the cathode. The two electrodes must be electrically connected to each other, allowing for a flow of electrons that leave the metal of the anode and flow through this connection to the ions at the surface of the cathode. This flow of electrons is an electrical current that can be used to do work, such as turn a motor or power a light.

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Explanation:

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3 0
2 years ago
What mass, in grams, of sodium bicarbonate, nahco3, is required to neutralize 1000.0 l of 0.350 m h2so4?
oksano4ka [1.4K]
Hello!

The net chemical equation between Sodium Bicarbonate and H₂SO₄ is the following one:

2NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2H₂O + 2CO₂

To calculate the amount of Sodium Bicarbonate needed to neutralize 1000 L of 0,350 M H₂SO₄ we'll need to use the following conversion factor, to go from the volume of H₂SO₄ to grams of Sodium Bicarbonate:

 1000 L H_2SO_4* \frac{0,350 moles H_2SO_4}{1 L}* \frac{2 moles NaHCO_3}{1 mol H_2SO_4}* \frac{84,007gNaHCO_3}{1molNaHCO_3} \\ \\ =58804,9 g NaHCO_3

So, to neutralize 1000 L  of 0,350 moles of H₂SO₄ you'll need 58804,9 grams of NaHCO₃
3 0
3 years ago
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