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Free_Kalibri [48]
3 years ago
15

Com relaçao a pilha mg/mg2+//cu2+/cu,indique

Chemistry
1 answer:
meriva3 years ago
6 0

Answer:

A: Cu; B: Mg; C: Mg + Cu²⁺ ⟶ Mg²⁺ + Cu

Explanation:

The left-hand side of the cell diagram is where oxidation occurs (the anode).

Mg ⟶ Mg²⁺ + 2e⁻

So, Mg is the anode.

The right-hand side of the cell diagram is where reduction occurs (the cathode)

Cu²⁺ + 2e⁻ ⟶ Cu

As the copper ions hit the copper cathode, they remove electrons from the metal and become copper atoms. Since the metal has lost electrons, Cu is the positive electrode.

To get the overall reaction, we add the two half-reactions,

Mg ⟶ Mg²⁺ + 2e⁻

<u>Cu²⁺ + 2e⁻ ⟶ Cu            </u>

Mg + Cu²⁺ ⟶ Mg²⁺ + Cu

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How to do part (b)??
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it has less tightly bound electrons, is able to lose electron easily as compare to metal B at it has 4 unpaired electron in 3d sub-shell.

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3 years ago
A sodium atom and sodium ion have the same
Ilia_Sergeevich [38]

<u>Answer:</u>

a) number of neutrons

<u>Explanation:</u>

A sodium ion is formed when an electron is removed from a sodium atom. This means that the atom's number of electrons changes, but the number of neutrons remains unchanged.

However, as the number of electrons changes, the electric charge and the electronic structure change, which means that a sodium atom and a sodium ion do not have the same number of electrons, nor do they have the same electric charge or electronic structure.

Therefore, option a) is the correct option.

3 0
2 years ago
Stars with a greater overall mass are and , and they consume their nuclear fuel .
kozerog [31]

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3 years ago
tea contains approximately 2% caffeine by weight. assuming that you started with 18g of tea leaves, calculate your percent yield
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4 0
3 years ago
Given the following data, determine the order of reaction with respect to H2.
alex41 [277]

Answer:

First-order with respect to hydrogen.

Explanation:

Hello!

In this case, considering that the rate law of this reaction can be expressed via:

r=k*p_{H_2}^m*p_{ICl}^n

If we want to know m, the order of reaction with respect to hydrogen, we need to relate the experiments 1 and 3 in order to get rid of the pressure of ICl:

\frac{r_1}{r_3} =\frac{k*p_{H_2,1}^m*p_{ICl,1}^n}{k*p_{H_2,3}^m*p_{ICl,3}^n}

Thus, we plug in the given rates, and pressures to get:

\frac{1.34}{0.266} =\frac{k*250^m*325^n}{k*50^m*325^n}

So we can cancel wout k and 325^n:

\frac{1.34}{0.266} =\frac{250^m}{50^m}

Next we solve for m, the order of reaction with respect to hydrogen:

5.04 =5^m\\\\log(5.04)=m*log(5)\\\\m=\frac{log(5.04)}{log(5)} \\\\m=1.0

It means it is first-order with respect to hydrogen.

Regards!

7 0
3 years ago
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