The concentration of AlCl3 solution if 150 ml of the solution contains 550 mg of cl- ion is 0.0344 M
calculation
concentration = moles /volume in liters
volume in liters = 150 /1000= 0.15 L
number of moles calculation
write the equation for dissociation of Al2Cl3
that is AlCl3 ⇔ Al^3+ + 3 Cl ^-
find the moles of Cl^- formed
moles =mass/molar mass
mass in grams= 550/ 1000 =0.55 grams
molar mass of Cl^- =35.5 g/mol
moles is therefore= 0.55/35.5 =0.0155 moles
by use of mole ration betweem AlCl3 to Cl^- which is 1:3 the moles of AlCl3 is =0.0155 x 1/3= 5.167 x10^-3 moles
concentration of AlCl3 is therefore= 5.167 x10^-3/ 0.15 =0.0344 M
Answer is: C) the fact that the number of lone pairs of electrons on the central atom is greater in the case of water.
Carbon(IV) oxide is nonpolar because CO₂ is linear molecule and the oxygen atoms are symmetrical (bond angles 180°).
Water is polar because of the bent shape of the molecule.
Oxygen atom in water molecule has sp3 hybridization. The bond angle between the two hydrogen atoms is approximately 104.45°.
Oxygen atom has atomic number 8, it means it has eight protons and eight electrons, so atom has neutral charge. Oxygen is a nonmetal.
Electron configuration of oxygen atom: ₈O 1s² 2s² 2p⁴.
Oxygen atom has six valence electrons
, two lone pairs and two electrons that form two sigma bonds with hydrogen atoms.
Carbon is a chemical element with symbol C and atomic number 6, which means it has 6 protons and six electrons. Four valence electrons are in 2s and 2p orbitals.
Electron configuration of carbon atom: ₆C 1s² 2s² 2p².
In carbon dioxide, carban has sp hybridization with no lone pairs.
Answer:
A)Chlorine and Bromine:
They are both non metal hence they form a covalent bond due to covalent bonding.
B)Potassium and Helium:
Helium ion has a small cationic radius and distorted by the potassium ion due to polarization.
C)Sodium and Lithium:
Both are metals hence they form a metallic bond since they share electrons to the electron cloud.
2.41 molecules, should be it
Three complete orders on each side of the m=0 order can be produced in addition to the m = 0 order.
The ruling separation is
d=1 / (470mm −1) = 2.1×10⁻³ mm
Diffraction lines occur at angles θ such that dsinθ=mλ, where λ is the wavelength and m is an integer.
Notice that for a given order, the line associated with a long wavelength is produced at a greater angle than the line associated with a shorter wavelength.
We take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.
That is, find the greatest integer value of m for which mλ<d.
since d / λ = 538×10⁻⁹m / 2.1×10 −6 m ≈ 3
that value is m=3.
There are three complete orders on each side of the m=0 order.
The second and third orders overlap.
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