Elements present in group 18 are known as noble gases. The outermost shell of these elements are completely filled.
18 is the answer
Answer is: a lower freezing point has solution of K₂SO₄.
Change in freezing
point from pure solvent to solution: ΔT =i · Kf · b.<span>
Kf - molal freezing-point depression constant for water is 1.86°C/m.
b - molality, moles of solute per
kilogram of solvent.
i - </span>Van't
Hoff factor.<span>
b(K</span>₂SO₄<span>) = 0.35 m.
</span>b(KCl) = 0.5 m.
i(K₂SO₄) = 3.
i(KCl) = 2.
ΔT(K₂SO₄) = 3 · 0.35 m · 1.86°C/m.
ΔT(K₂SO₄) = 1.953°C.
ΔT(KCl) = 2 · 0.5 m · 1.86°C/m.
ΔT(KCl) = 1.86°C.
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:
The mass of xenon in the compound is 2.950 grams
Explanation:
Step 1: Data given
Mass of XeF4 = 4.658 grams
Molar mass of XeF4 = 207.28 g/mol
Step 2: Calculate moles of XeF4
Moles XeF4 = mass XeF4 / molar mass XeF4
Moles XeF4 = 4.658 grams / 207.28 g/mol
Moles XeF4 = 0.02247 moles
Step 3: Calculate moles of xenon
XeF4 → Xe + 4F-
For 1 mol xenon tetrafluoride, we have 1 mol of xenon
For 0.02247 moles XeF4 we have 0.02247 moles Xe
Step 4: Calculate mass of xenon
Mass xenon = moles xenon * molar mass xenon
Mass xenon = 0.02247 moles * 131.29 g/mol
Mass xenon = 2.950 grams
The mass of xenon in the compound is 2.950 grams
Answer:
a. the maximum number of σ bonds that the atom can form is 4
b. the maximum number of p-p bonds that the atom can form is 2
Explanation:
Hybridization is the mixing of at least two nonequivalent orbitals, in this case, we have the mixing of one <em>s, 3 p </em> and <em> 2 d </em> orbitals. In hybridization the number of hybrid orbitals generated is equal to the number of pure atomic orbital, so we have 6 hybrid orbital.
The shape of this hybrid orbital is octahedral (look the attached image) , it has 4 orbital located in the plane and 2 orbital perpendicular to it.
This shape allows the formation of maximum 4 σ bond, because σ bonds are formed by orbitals overlapping end to end.
And maximum 2 p-p bonds, because p-p bonds are formed by sideways overlapping orbitals. The atom can form one with each one of the orbitals located perpendicular to the plane.
