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PilotLPTM [1.2K]
3 years ago
5

Explain in words or using your diagrams how beryllium atoms would react with fluorine atoms.

Chemistry
1 answer:
Marysya12 [62]3 years ago
8 0

Explanation:

To delineate the the nature of the bonds that would be formed between the two elements, let us first write the electronic configuration of the two species;

  Be  = 2, 2

  F  = 2, 7

 Beryllium is a metal with two valence electrons whereas fluorine is a halogen with seven valence electrons.

When Be loses two electrons it becomes isoelectronic with He;

                 Be →  Be²⁺ + 2e⁻  

Also, when fluorine gains  an electron, it becomes isoelectronic with Ne;

               F  + e⁻ →  F⁻

This loss and gain of electrons between the two elements creates an electrostatic attraction them and they enter into an electrovalent bond.

      Hence;

           Be²⁺  + 2F⁻ → BeF₂

You might be interested in
Plz help guys ASAP! <br> Thanks in advance
MariettaO [177]

Answer:

3.6 moles

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

1 mole of N₂ reacted with 3 moles of H₂ to produce 2 moles NH₃.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N₂ reacted with 3 moles of H₂.

Therefore, 3.2 moles of N₂ will react with = 3.2 × 3 = 9.6 moles of H₂.

From the calculation made above, we can see that it will take a higher amount (i.e 9.6 moles) of H₂ than what was given (i.e 5.4 moles) to react completely with 3.2 moles of N₂.

Therefore, H₂ is the limiting reactant and N₂ is the excess reactant.

Finally, we the greatest quantity of ammonia, NH₃ produced from the reaction.

In this case, the limiting reactant will be use because all of it is consumed in the reaction.

The limiting reactant is H₂ and greatest quantity of ammonia, NH₃ produced can be obtained as follow:

From the balanced equation above,

3 moles of H₂ reacted to produce 2 moles NH₃.

Therefore, 5.4 of H₂ will react to produce = (5.4 × 2)/3 = 3.6 moles of NH₃

Thus, the greatest quantity of ammonia, NH₃ produced from the reaction is 3.6 moles

4 0
3 years ago
What is first and second element of the periodic table?
stira [4]
The elements on the periodic table are listed in increasing atomic number. 

Hydrogen is the first element, and has an A.N. of 1. Also, its very interesting how it doesn't need 8 valence electrons to be stable. 

The second element is Helium, which has an A.N (atomic number) of two. 
7 0
3 years ago
Read 2 more answers
What is the fourth largest element on the periodic table?
KatRina [158]
I think it's the Pentium up
 115 
<span>he's in the 15th column, 7th period </span><span />
4 0
3 years ago
Read 2 more answers
An element crystallizes in a face-centered cubic lattice. If the length of an edge of the unit cell is 0.408 nm, and the density
V125BC [204]

Answer:

Au

Explanation:

For the density of a face-centered cubic:

Density = \dfrac{4 \times M_w}{N_A \times a^3}

where

M_w = molar mass of the compound

N_A= avogadro's constant

a^3 = the volume of a unit cell

Given that:

Density (\rho) = 19.30 g/cm³

a = 0.408 nm

a = 0.408 \times 10^{-9} \times 10^{2} \ cm

a = 4.08 \times 10^ {-8} \ cm

∴

19.3 = \dfrac{4 \times M_w}{(6.023 \tmes 10^{23})\times (4.08 \times 10^{-8})^3}

M_w= \dfrac{19.3\times (6.023 \times 10^{23})\times (4.08 \times 10^{-8})^3}{4}

M_w=197.37 \ g/mol

Thus, the molar mass of 197.37 g/mol element is Gold (Au).

4 0
3 years ago
Purification of nickel can be achieved by electrorefining nickel from an impure nickel anode onto a pure nickel cathode in an el
Alexxandr [17]

Answer: 530 hours

Explanation:

The reduction of Nickel ions to nickel is shown as:

Ni^{2+}+2e^-\rightarrow Ni

96500\times 2=193000Coloumb of electricity deposits 1 mole of Nickel

1 mole of Nickel weighs = 58.7 g

Given quantity = 18.0 kg = 18000 g  (1kg=1000g)

58.7 g of Nickel is deposited by 193000 C of electricity

18000 g of Nickel is deposited by = \frac{193000}{58.7}\times 18000=59182282.8C of electricity

Q=I\times t

where Q= quantity of electricity in coloumbs  = 59182282.8C

I = current in amperes = 31.0 A

t= time in seconds = ?

59182282.8C=31.0A\times t

t=1909105.9sec

(1h=3600 sec)

t=530hours

Thus 530 hours are required to plate 18.0 kg of nickel onto the cathode if the current passed through the cell is held constant at 31.0 A

3 0
3 years ago
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