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german
3 years ago
15

Which of the following particles are free to drift in metals?

Chemistry
1 answer:
Nostrana [21]3 years ago
8 0

Answer:

Electrons

Explanation:

"This free-floating collection of electrons gives metals many of their unique properties, and is sometimes called a “sea of electrons,”

Learn more about it at this website:

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=2ahUKEwjUw9Sbl7nmAhWQKs0KHXFGCakQFjACegQICxAK&url=https%3A%2F%2Futexaslearn.instructure.com%2Fcourses%2F2010099%2Fpages%2Fexplain-i-metallic-bonding-and-properties-2&usg=AOvVaw3B0fp0MY7ut9ssl_ebAgP7

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Find the mass of 3.02 mol Cl2.<br> Answer in units of g.
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Explanation:

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Free-energy change, ΔG∘, is related to cell potential, E∘, by the equation ΔG∘=−nFE∘ where n is the number of moles of electrons
elena-s [515]

Answer:

-372000 J or -372 KJ

Explanation:

We have the electrochemical reaction as;

Mg(s)  +  Fe^2+(aq)→  Mg^2+(aq)  +   Fe(s)

We must first calculate the E∘cell from;

E∘cathode -  E∘anode

E∘cathode = -0.44 V

E∘anode = -2.37 V

Hence;

E∘cell = -0.44 V -(-2.37 V)

E∘cell = 1.93 V

n= 2 since two electrons were transferred

F=96,500C/(mol e−)

ΔG∘=−nFE∘

ΔG∘= -( 2 * 96,500 * 1.93)

ΔG∘= -372000 J or -372 KJ

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A flexible container at an initial volume of 4.11 L contains 2.51 mol of gas. More gas is then added to the container until it r
Nitella [24]

Answer:

7.81 moles

Explanation:

To solve this problem, let us generate an expression involving volume and number of mole of the gas since the pressure and temperature of the gas are constant.

From ideal gas equation:

PV = nRT

Divide both side by P

V= nRT/P

Divide both side by n

V/n = RT/P

Since RT/P are constant, then:

V1/n1 = V2/n2

Data obtained from the question include:

V1 = 4.11

n1 = 2.51 moles

V2 = 16.9L

n2 =?

Using the above equation i.e V1/n1 = V2/n2, the final number of the gas can be obtained as illustrated below:

4.11/2.51 = 16.9/n2

Cross multiply to express in linear form

4.11 x n2 = 2.51 x 16.9

Divide both side by 4.11

n2 = (2.51 x 16.9) / 4.11

n2 = 10.32moles

Now, to obtain the number of mole of the gas added, we'll subtract the initial mole from the final mole i.e

n2 — n1

Number of mole added = n2 — n1

10.32 — 2.51 = 7.81 moles

Therefore, 7.81 moles of the gas was added to the container

3 0
3 years ago
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