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Agata [3.3K]
3 years ago
12

Why does common table salt (NaCl) have a high melting point?

Chemistry
1 answer:
schepotkina [342]3 years ago
8 0
The correct answer is c
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An elements atomic number tells you the number of
liq [111]
The number of protons in the nucleus of the atom, I believe.  c:
8 0
4 years ago
What word or two-word phrase best describes the shape of the fluoroform molecule?
Vlad [161]

Answer : The shape of the fluoroform molecule is Tetrahedral.

Explanation :

First we have to calculate the Hybridization of the molecule by formula,

\text{Number of electrons} = \frac{1}{2}[V+H-C+A]

where,

V = Number of valence shell electron in central atom

H = Number of neighboring monovalent atom

C = charge of cation

A = charge of anion

The central atom in this molecule is Carbon, it has 4 electrons in their valence shell.

The neighboring monovalent atoms are one Hydrogen atom and three Fluorine atom.

There is no charge of cation and anion on the given molecule.

V = 4

H = 1 Hydrogen atom + 3 fluorine atom = 4

C = 0

A = 0

By the above hybridization formula, we get

\text{Number of electrons}= \frac{1}{2}[4+4-0+0] = 4

The number of electron pair = 4

The number of lone pair = 0

The number of electrons is 4, this means that the hybridization is sp^{3} and the geometry of the molecule is Tetrahedral.

The geometry of the molecule is shown below.


3 0
3 years ago
The energy of any one-electron species in its nth state (n = principal quantum number) is given by E = –BZ2 /n2 where Z is the c
Ivahew [28]

Explanation:

(a) The given data is as follows.

            B = 2.180 \times 10^{-18} J

            Z = 4 for Be

Now, for the first excited state n_{f} = 2; and n_{i} = \infinity if it is ionized.

Therefore, ionization energy will be calculated as follows.

         I.E = \frac{-Bz^{2}}{\infinity^{2}} - (\frac{-2.180 \times 10^{-18} J /times (4)^{2}}{(2)^{2}})

              = 8.72 \times 10^{-18} J

Converting this energy into kJ/mol as follows.

           8.72 \times 10^{-18} J \times 6.02 \times 10^{23} mol  

           = 5249 kJ/mol

Therefore, the ionization energy of the Be^{3+} ion in its first excited state in kilojoules per mole is 5249 kJ/mol.

(b) Change in ionization energy is as follows.

         \Delta E = -Bz^{2}(\frac{1}{(4)^{2}} - {1}{(2)^{2}}) = \frac{hc}{\lambda}

   \frac{hc}{\lambda} = 0.1875 \times 2.180 \times 10^{-18} J \times (4)^{2}                

        \lambda = \frac{6.626 \times 10^{-34} \times 2.998 \times 10^{8} m/s}{0.1875 \times 2.180 \times 10^{-18} J \times 16}

                     = 303.7 \times 10^{-10} m

or,                 = 303.7^{o}A

Therefore, wavelength of light given off from the Be^{3+} ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels 303.7^{o}A.

5 0
3 years ago
Using the HH eq. calculate the amounts (masses) of sodium phosphate monobasic monohydrate (NaH2PO4 H2O) and sodium phosphate dib
pishuonlain [190]
Drop out my dude whatever this is ain’t worth it
4 0
3 years ago
In the periodic table of elements, some elements share similar properties and have the same number of valence electrons in their
Lilit [14]
The type of element is group
5 0
3 years ago
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