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n200080 [17]
1 year ago
14

discuss how variations in electronegativity result in the unequal sharing of electrons in polar molecules.

Chemistry
1 answer:
konstantin123 [22]1 year ago
6 0

Variations in electronegativity prompt in the unequal halves of electrons in polar molecules because when one atom is more electronegative than the other, it becomes more polar than the other.

It results in the more electronegative atom to have a slightly negative (-ve) charges, and the other atom to have partial or slightly positive(+ve) charges.

Polar molecules have unequal sharing of electrons because the atoms have unequal attraction for electrons so the sharing is unequal.

The larger the difference in electronegativity between the two atoms, the more the polar the bond.

Hydrogen bonds are involved in unequal sharing of electrons between two atoms.

To know more about variations in electronegativity in polar molecules here :

brainly.com/question/18260584?referrer=searchResults

#SPJ4

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If the [H3O+] of a solution is 1x 10-8 mol/L the [OH-] is
Studentka2010 [4]

Answer:

<em>(H30+)= 1x10^-6 M</em>

Explanation:

Both pH and pOH have a relationship to belonging to the same aqueous solution: the expression of the Kwater (ionic product of the water Kw) is used:

1x 10-8 mol/L equals to1x10-8 M

(H3O+) x (OH-) = 1x10^-14

(H30+)x 1x 10^-8 =1x10^-14

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<em>(H30+)= 1x10^-6 M</em>

6 0
3 years ago
A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat
Lady_Fox [76]

Explanation:

(a)  The given data is as follows.  

         Load applied (P) = 1000 kg

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Expression for Brinell Hardness is as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}    

Now, putting the given values into the above formula as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}                            = \frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}  

                       = \frac{2000}{9.98}                          

                       = 200

Therefore, the Brinell HArdness is 200.

(b)     The given data is as follows.

               Brinell Hardness = 300

                Load (P) = 500 kg

               BHI diameter (D) = 10 mm

             Indentation produced (d) = ?

                      d = \sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}

                         = \sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}

                          = 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

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