) A concentration cell is based on the aqueous reaction

Chemistry

1 answer:

Brrunno [24]4 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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How many grams of CaCl2 are in 250 mL of 2.0 M CaCl2?

Tom [10]

The answer is: " 56 g CaCl₂ " .
__________________________________________________________

Explanation:
__________________________________________________________
2.0 M CaCl₂ = 2.0 mol CaCl₂ / L ;

Since: "M" = "Molarity" (measurement of concentration);

= moles of solute per L {"Liter"} of solution.
__________________________________________________________
Note the exact conversion: 1000 mL = 1 L .

Given: 250 mL ;

250 mL = ? L ? ;

250 mL * (1 L / 1000 L) = (250/1000) L = 0.25 L .
___________________________________________________________

(2.0 mol CaCl₂ / L ) * (0.25L) = (2.0) * (0.25) mol = 0.50 mol CaCl₂ ;

We have: 0.50 mol CaCl₂ ; Convert to "g" (grams):

→ 0.50 mol CaCl₂ .
___________________________________________________________
1 mol CaCl₂ = ? g ?

From the Periodic Table of Elements:

1 mol Ca = 40.08 g

1 mol Cl = 35.45 g .

There are 2 atoms of Cl in " CaCl₂ " ;

→ Note the subscript, "2", in the " Cl₂ " ;
__________________________________________________________
So, to calculate the molar mass of "CaCl₂" :

40.08 g + 2(35.45 g) =

40.08 g + 70.90 g = 110.98 g ; round to 4 significant figures;

→ round to 111 g/mol .
__________________________________________________________
So:

→ 0.50 mol CaCl₂ = ? g CaCl₂ ? ;

→ 0.50 mol CaCl₂ * (111 g CaCl₂ / mol CaCl₂) ;

= (0.50) * (111 g) CaCl₂ ;

= 55.5 g CaCl₂ ;

 → round to 2 significant figures;

 → 56 g CaCl₂ .
___________________________________________________________
The answer is: " 56 g CaCl₂ " .
___________________________________________________________

6 0

3 years ago

Read 2 more answers

11. What is the maximum number of electrons that an atomic orbital can contain?

Arada [10]

Answer:

Explanation:

Each orbital can hold two electrons. One spin-up and one spin-down.

6 0

3 years ago

If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming tha

NARA [144]

There are 2.32 x 10^6 kg sulfuric acid in the rainfall.

Solution:
We can find the volume of the solution by the product of 1.00 in and 1800 miles2:
1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m
1.00 in * 1 m / 39.3701 in = 0.0254 m
Volume = 4.662 x 10^9 m^2 * 0.0254 m
= 1.184 x 10^8 m^3 * 1000 L / 1 m3
= 1.184 x 10^11 Liters

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70:
[H+] = 10^-pH = 10^-3.7 = 0.000200 M
[H2SO4] = 0.000100 M

By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid:
1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4

We can now calculate for the mass of sulfuric acid in the rainfall:
mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
= 2.32 x 10^9 g * 1 kg / 1000 g
= 2.32 x 10^6 kg H2SO4

3 0

4 years ago

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A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a

svlad2 [7]

Answer: The amount of oxygen gas consumed by mouse is 0.202 grams.

Explanation:

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows: