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geniusboy [140]
3 years ago
8

Bonds between two atoms that are equally electronegative are _____. bonds between two atoms that are equally electronegative are

_____. ionic bonds hydrogen bonds nonpolar covalent bonds polar covalent bonds
Chemistry
2 answers:
kykrilka [37]3 years ago
8 0

Answer:

Non-polar covalent  bond.

Explanation:

Ionic bond:It is that type of bond which is formed by gain or loss of electrons between two atoms.

Hydrogen bond:It is that type of bond in which hydrogen atom is linked with atom of high electronegativity.

Polar covalent bond: It is formed by sharing of electrons of electrons between two atoms of different electronegativity.

Non-Polar covalent bond:It is formed by sharing of electrons between two atoms of same electronegativity.

Bond between two atoms that are equally  electronegative are non-polar covalent bond.

DanielleElmas [232]3 years ago
7 0
Bonds between two atoms that are equally electronegative are nonpolar covalent bonds
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A pressure of 745 mm hg equals select one:
Flura [38]
As we know that,

                                        1 mm Hg  =  1 torr
So,
                                    745 mm Hg  =  X

                         X  =  (745 mm Hg × 1 torr) ÷ 1 mm Hg

                         X  =  745 torr

Also,
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So,
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3 0
3 years ago
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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