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Crazy boy [7]
3 years ago
10

1.Explain why the reactivity of group 7 decreases as you move down the group. Try to use the sentence starters here: When group

7 elements react, their atoms….
As you move down the group, the outer shells get…

Therefore the force of attraction between the shells and the nucleus is….

This makes attracting an extra electron…

Consequently, reactivity…​
Chemistry
1 answer:
Kruka [31]3 years ago
5 0

Answer: The reactivity of group 7 decreases as we move down the group because:

Explanation:

The elements of group 7 that is fluorine to iodine. The halogens are non metals and they react with metals to gain electrons. The metals loose electrons and the non metal gains it.

As we move down the group  the atomic radius gets bigger( more electron and more proton) and as a result the outer shells move further away from the nucleus.

There is more distance between the negatively charged electrons and positively charged nucleus.

Therefore the force of attraction between the shells and nucleus is lesser or weaker.

This makes attracting an extra electron from metals very difficult which results in weaker reaction.

Consequently, the reactivity decreases as we move down the group 7

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How many moles of a gas would occupy 11.4 L at 273K and 2.00 atm?
bagirrra123 [75]

Answer:

1.02mol

Explanation:

Using the general gas equation below;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the information provided in this question,

P = 2.0 atm

V = 11.4L

T = 273K

n = ?

Using PV = nRT

n = PV/RT

n = 2 × 11.4/ 0.0821 × 273

n = 22.8/22.41

n = 1.017

n = 1.02mol

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3 years ago
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3 years ago
For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [
sveta [45]

Explanation:

The given data is as follows.

     E^{o} = 0.483,     [Co^{3+}] = 0.173 M,

     [Co^{2+}] = 0.433 M,     [Cl^{-}] = 0.306 M,

     P_{Cl_{2}} = 9.0 atm

According to the ideal gas equation, PV = nRT

or,             P = \frac{n}{V}RT    

Also, we know that

                Density = \frac{mass}{volume}

So,         P = MRT

and,          M = \frac{P}{RT}

                    = \frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}

                    = \frac{9.0}{24.436}

                    = 0.368 mol/L

Now, we will calculate the cell potential as follows.

          E = E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}

             = 0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}

             = 0.483 - 0.02955 log \frac{0.0689}{0.0162}

             = 0.483 - 0.02955 \times 0.628

             =  0.483 - 0.0185

             = 0.4645 V

Thus, we can conclude that the cell potential of given cell at 25^{o}C is 0.4645 V.

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Answer:

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