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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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15. If you dilute a 6 M solution of HCl from 5 mL to 50mL, what is the concentration of this new solution? (M1V1 = M2V2)
Andreyy89

Answer:

B) 0.6M

Explanation:

               I apologize in advance if it is not correct :l

The (M1V1= M2V2) is given for you to plug in the correct numbers so let's jot this down.

                    (M1*V1= M2*V2)

so they give us 6M which would be our (M1), from this we can also conclude that 5mL is also V1; ( if you notice the M1's and V1's are always found next to eachother). This leads us to our 50mL, this would be our V2 because the volume went from 5mL to 50mL. Now lets put this in order based on what we know.

M1= 6M                                (M1*V1= M2*V2)

V1= 5mL

M2= ?

V2= 50mL

now we plug in what we know into the equation to find the unknown (M2)

                   (6M*5mL= M2*50mL)

now we could do the long math, but I don't think your on brainly to do the hard way. so lets keep it simple!

                   We are going to put the 50mL under the (6M*5mL) for division.

                             \frac{(6M*5mL)}{(50mL)} This is honestly MUCH easier, than manually answering. you just put that in the calculator and it'll give you B) 0.6M

                 

   honestly though I might not know what I'm doing cuz im currently doing my test and decided to answer this question ;)

        Good Luck!

6 0
3 years ago
If a compound begins with a metal, it most likely is a _______ compound
Bess [88]

Answer:

Metallic compound

Explanation:

Metallic compound -

A metallic compound refers to the species having at least one metal, is known  as a metallic compound.

In general a metallic compound consists of two ions, a positive ion or the cation, is is basically a metal, and,

a negative ion or an anion, which is a non - metal.

Hence, from the given statement of the question,

The correct answer is - metallic compound.  

8 0
3 years ago
in an experiment 3.425g of lead oxide was reduced to form 3.105g of lead the empirical formula of the lead oxide is?​
Romashka-Z-Leto [24]

Answer:

Pb3O4

Explanation:

According to this question, 3.425g of lead oxide was reduced to form 3.105g of lead in an experiment. Since lead oxide contains both lead (Pb) and oxygen (O) element,

Mass of lead oxide = 3.425g

Mass of lead = 3.105g

Mass of oxygen = (3.425g - 3.105g) = 0.320g

Next, we convert each mass value to mole by dividing by respective molar mass

Pb = 3.105g ÷ 207.2 = 0.0149mol

O = 0.320g ÷ 16 = 0.02mol

Next, we divide each mole value by the smallest (0.0149)

Pb = 0.0149mol ÷ 0.0149mol = 1

O = 0.02mol ÷ 0.0149mol = 1.342

Multiply each ratio value by 3 to get:

Pb = 1 × 3 = 3

O = 1.342 × 3 = 4.026

The whole number ratio, approximately, of Pb and O is 3:4, hence, their empirical formula is Pb3O4.

4 0
2 years ago
50 POINTS! ANSWER FAST!
Korvikt [17]

<u>Given:</u>

The initial energy of the electron Einitial = 16.32 * 10⁻¹⁹ J

The energy released i.e the change in energy ΔE = 5.4 * 10⁻¹⁹ J

<u>To determine:</u>

The final energy state Efinal of the electron

<u>Explanation:</u>

Since energy is being released, this suggests that Efinal < Einitial

i.e. ΔE = Einitial - Efinal

Efinal = Einitial - ΔE = (16.32 - 5.4)*10⁻¹⁹ = 10.92 * 10⁻¹⁹ J

Ans: A)

The electron moved down to an energy level and has an energy of 10.92 * 10⁻¹⁹ J

6 0
3 years ago
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3 years ago
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