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Sever21 [200]
3 years ago
6

) A concentration cell is based on the aqueous reaction

Chemistry
1 answer:
Brrunno [24]3 years ago
8 0

Answer:

Kindly check the attachment for the diagram representing the cell and showing all necessary components including: anode, cathode, electron flow, cation flow and anion flow.

Explanation:

So, the reaction in the concentration cell is given below as;

Cu2+(1.00 M) → Cu2+(0.0100 M).--(1).

The anode = Cu2+(0.0100 M) because it has lesser Concentration, thus, lesser potential value.

Cathode = Cu2+(1.00 M) because it has higher Concentration, hence higher potential value.

It must be noted that in the digaram depicting the Reaction, the electrons moves from the anode part of the cell to the cathode part of the cell and this is done through an external circuit. The following are the things that happens at each electrode;

At the Anode: in here is where oxidation occurs and Cu^2+ is released into the solution.

At the cathode: in here is where the reduction occur and the Cu^2+ moves in the direction to where the Cu electrode is, thus, causing the deposition of Cu.

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2 years ago
Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:
nignag [31]

Answer:  1) Maximum mass of ammonia  198.57g  

2) The element that would be completely consumed is the N2

3) Mass that would keep unremained, is the one of  the excess Reactant, that means the H2 with 3,44g

Explanation:

  • In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:

N2(g) + 3H2(g) ⟶2NH3(g)

Both equal amount of atoms side to side.

  • Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the  molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)

Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3

Moles of ammonia produced with 38.77 g H2⟶  38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3

  • As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element  that would be completey consumed, and the maximum mass of ammonia will be produced from it.
  • We proceed calculating the maximum mass of NH3 from the 163.3g of N2.

11.66  mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3

  • In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:

163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2

That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of  H2, thus, 38.77g - 35.33 g = 3.44g H2 is left

3 0
3 years ago
(pls help)
WITCHER [35]
Anthony’s because it explains way more
8 0
2 years ago
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