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tigry1 [53]
4 years ago
10

where are the most reactive elements on the periodic table ? what makes one element more reactive than another​

Chemistry
1 answer:
DedPeter [7]4 years ago
7 0

Lower left corner and the upper right corner before the noble gases

Number of valence electrons

Explanation:

The most reactive elements are the most electropositive and electronegative elements.

Francium is the most reactive metal and flourine the most reactive non-metals.

Francium is the most electropositive metal and fluorine is the most electronegative non-metal.

Electropositivity is the ability of an atom to willingly lose its valence electrons. This is typical of metals.

Electronegativity is the ability of atoms to attract the shared electrons to themselves in a bond.

  • An element is more reactive than another based on the number of valence electrons it has.
  • The valence electrons determines how reactive an atom is or would be.
  • Elements that requires just one or two electrons to complete their octet would be highly reactive and dire need to find such electron.
  • Also, elements with just an electron above the stable configuration will easily want to lose it.
  • Elements with a complete configuration will  not be reactive.

The valence electrons are the electrons in the outer most shell of an atom used in chemical bonding.

learn more:

Reactivity brainly.com/question/6496202

#learnwithBrainly

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Answer:

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4 years ago
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In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,
Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -              -

At t =equilibrium      (x-s)                s           2s          

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

K_{sp}=s\times {2s}^2

K_{sp}=4s^3

Given  s = 7.4×10⁻³ M

So, Ksp is:

K_{sp}=4\times (7.4\times 10^{-3})^3

K_{sp}=4\times (7.4\times 10^{-3})^3

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -            0.20

At t =equilibrium      (x-s')             s'         0.20+2s'         

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2

Solving for s', we get

<u>s' = 4.0×10⁻⁵ M</u>

<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.</u>

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Learn more:

Biological catalysts brainly.com/question/8413755

#learnwithbrainly

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