Dipole-dipole interactions, and London dispersion interactions
First solve the moles of oxgen present in the compound
mol O = 6.93 g O ( 1 mol O / 16 g O )
mol O = 0.43 mol H
then solve the moles of hydrogen present
mol H = ( 7.36 - 6.93) g H ( 1 mol H / 1 g H)
mol H = 0.43 mol H
so the O and H are in the same mole content so the molecular formula would be OH, but the molar mass will not satisfy. so the answer would be
H2O2
Answer: 2.48×10^-17 J
Explanation:
Given the following :
Wavelength = 8nm (8 x 10^-9 m)
Energy(e) of X-ray =?
Energy=[speed of light(c) × planck's constant (h)] ÷ wavelength
Speed of light = 3×10^8m/s
Planck's constant = 6.626×10^-34 Js
Wavelength = 8 x 10^-9 m
Energy = [(3×10^8) * (6.626×10^-34)] / 8 x 10^-9
Energy = [19.878×10^(8-34)] / 8 x 10^-9
Energy = 2.48475 × 10^(-26+9)
Energy = 2.48×10^-17 J
Answer:
Elemental gold to have a Face-centered cubic structure.
Explanation:
From the information given:
Radius of gold = 144 pm
Its density = 19.32 g/cm³
Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:
![a = \sqrt{8} r](https://tex.z-dn.net/?f=a%20%3D%20%5Csqrt%7B8%7D%20r)
![a = \sqrt{8} \times 144 pm](https://tex.z-dn.net/?f=a%20%3D%20%5Csqrt%7B8%7D%20%5Ctimes%20144%20pm)
a = 407 pm
In a unit cell, Volume (V) = a³
V = (407 pm)³
V = 6.74 × 10⁷ pm³
V = 6.74 × 10⁻²³ cm³
Recall that:
Net no. of an atom in an FCC unit cell = 4
Thus;
![density = \dfrac{mass}{volume}](https://tex.z-dn.net/?f=density%20%3D%20%5Cdfrac%7Bmass%7D%7Bvolume%7D)
![density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}](https://tex.z-dn.net/?f=density%20%3D%20%5Cdfrac%7B%204%20%5C%20atm%20%28%20196.97%20%5C%20g%2Fmol%29%20%28%5Cdfrac%7B1%20%5C%20mol%20%7D%7B6.022%20%5Ctimes%2010%5E%7B23%7D%20%5C%20atoms%7D%29%7D%7B6.74%20%5Ctimes%2010%5E%7B-23%7D%20%5C%20cm%5E3%7D)
density d = 19.41 g/cm³
Similarly; For a body-centered cubic structure
![r = \dfrac{\sqrt{3}}{4}a](https://tex.z-dn.net/?f=r%20%3D%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B4%7Da)
where;
r = 144
![144 = \dfrac{\sqrt{3}}{4}a](https://tex.z-dn.net/?f=144%20%3D%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B4%7Da)
![a = \dfrac{144 \times 4}{\sqrt{3}}](https://tex.z-dn.net/?f=a%20%3D%20%5Cdfrac%7B144%20%5Ctimes%204%7D%7B%5Csqrt%7B3%7D%7D)
a = 332.56 pm
In a unit cell, Volume V = a³
V = (332.56 pm)³
V = 3.68 × 10⁷ pm³
V 3.68 × 10⁻²³ cm³
Recall that:
Net no. of atoms in BCC cell = 2
∴
![density = \dfrac{mass}{volume}](https://tex.z-dn.net/?f=density%20%3D%20%5Cdfrac%7Bmass%7D%7Bvolume%7D)
![density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}](https://tex.z-dn.net/?f=density%20%3D%20%5Cdfrac%7B%202%20%5C%20atm%20%28%20196.97%20%5C%20g%2Fmol%29%20%28%5Cdfrac%7B1%20%5C%20mol%20%7D%7B6.022%20%5Ctimes%2010%5E%7B23%7D%20%5C%20atoms%7D%29%7D%7B3.68%20%5Ctimes%2010%5E%7B-23%7D%20%5C%20cm%5E3%7D)
density =17.78 g/cm³
From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.
This makes the elemental gold to have a Face-centered cubic structure.
Answer:
ionic compounds are made from metal and non mental elements
Explanation:
in this case it is know as an ionic compound because it contains a charge. For example, NaCl is simply a compound as it contains no charges (the charges cancel out as Cl is -1 and Na is +1)
but OH- is an ionic compound as it has a charge if -1 (O has a -2 charge and H has a +1 charge, so -2+1=-1 so OH has -1 charge)