Answer:
This would support Dalton's postulates that proposed the atoms are indivisible because no small particles are involved.
Explanation:
Experiment using the gas discharge tube by J.J Thomson led to the discovery of cathode rays which are now known as electrons.
Primarily, Thomson's experiment led to the discovery of cathode rays, electrons, as subatomic particles.
If the size of the atoms observed at the cathode is the same as that of the rays,we can conclude that the particles of the rays are the simplest form of matter we can have. This would suggest that the atom is indeed the smallest indivisible particle of a matter according to Dalton.
<span>Electrons in a nitrogen-phosphorus covalent bond are not shared equally because nitrogen and phosphorus do not have the same electronegativity. The atoms spend more time around the most electronegative atom nitrogen.</span>
This is an incomplete question, the given sketch is shown below.
Answer : The name of given unit cell is, FCC (face-centered cubic unit cell)
Explanation :
Unit cell : It is defined as the smallest 3-dimensional portion of a complete space lattice which when repeated over the and again in different directions produces the complete space lattice.
There are three types of unit cell.
- SCC (simple-centered cubic unit cell)
- BCC (body-centered cubic unit cell)
- FCC (face-centered cubic unit cell)
In SCC, the atoms are arranged at the corners.
The number of atoms of unit cell = Z = 1
In BCC, the atoms are arranged at the corners and the body center.
The number of atoms of unit cell = Z = 2
The given unit cell is, FCC because the atoms are arranged at the corners and the center of the 6 faces.
The number of atoms of unit cell = Z = 4
Thus, the name of given unit cell is, FCC (face-centered cubic unit cell)
<span>Avogadro's number
represents the number of units in one mole of any substance. This has the value
of 6.022 x 10^23 units / mole. This number can be used to convert the number of
atoms or molecules into number of moles. We calculate as follows:
</span>1.40x10^23 molecules of N2 ( 1 mol / 6.022 x 10^23 molecules ) ( 28.02 g / mol ) = 6.51 g N2
Ksp of AgCl= 1.6×10⁻¹⁰
AgCl=Ag⁺ +Cl⁻
Ksp=[Ag⁺][Cl⁻]
Assume [Ag⁺]=[Cl⁻]=x
Ksp=x²
1.6×10⁻¹⁰=x²
x=0.000012
In FeCl₃:
FeCl₃------>Fe⁺³+ 3Cl⁻
as there is 0.010 M FeCl₃
So there will be ,
[Cl⁻]= 0.030
So
[Ag⁺]=Ksp/[Cl⁻]
=1.6×10⁻¹⁰/0.030
=5.3×10⁻⁹
so solubility of AgCl in FeCl₃ will be 5.3×10⁻⁹.
