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rusak2 [61]
2 years ago
8

When two fluorine atoms bond together in F2, what type of covalent bond do they form?

Chemistry
2 answers:
andrew11 [14]2 years ago
5 0

Answer:

A single bond bc they overlap to share ONE pair of electrons

Explanation:

Took the test

makvit [3.9K]2 years ago
4 0

When two  fluorine  atoms  bond  together  in f2  the  type of covalent  bond   do they  form is non-polar covalent   bond.


  Explanation

They are  two type of covalent bond.  that is non polar covalent bond and  polar covalent  bond.


Non polar  covalent  bond  is formed  when atoms share  a pair  of electrons  with each other  equally.

polar covalent  bond is bond formed when  atoms share  a pair   electrons unequally.


 F2  is a non polar  since  it is made  up of two  same atoms   which  has  the  same electronegativity, therefore equal number of electrons exist  in the  orbital overlap.


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Identify the intermolecular attractions for dimethyl ether and for ethyl alcohol. Which molecule is expected to be more soluble
zheka24 [161]

Answer:

See explanation

Explanation:

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So, the only kind of intermolecular interaction that exists in dimethyl ether is London dispersion forces.

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3 0
3 years ago
1.65g of zinc is used to make 8g of zinc iodide. How much iodine is required for this reaction?
creativ13 [48]

Answer:

6.45 g of iodine, I₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Zn + I₂ —> ZnI₂

Next, we shall determine the mass of Zn and I₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of Zn = 65 g/mol

Mass of Zn from the balanced equation = 1 × 65 = 65 g

Molar mass of I₂ = 127 × 2 = 254 g/mol

Mass of I₂ from the balanced equation = 1 × 254 = 254 g

SUMMARY:

From the balanced equation above,

65 g of Zn reacted with 254g of I₂.

Finally, we shall determine the mass of f I₂ needed to react with 1.65 g of Zn. This can be obtained as follow:

From the balanced equation above,

65 g of Zn reacted with 254g of I₂.

Therefore, 1.65 g of Zn will react with = (1.65 × 254)/65 = 6.45 g of I₂.

Thus, 6.45 g of iodine, I₂ is needed for the reaction.

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