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

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