Answer:
Option c, Two atomic orbitals combine to form one molecular orbital
Explanation:
Molecular orbitals are formed by linear combination of atomic orbitals.
Some of the important facts of molecular orbital theories are as follows:
- No. of the molecular orbitals formed are equal to the no. of atomic orbitals participated.
- Half of the molecular orbitals are bonding molecular orbitals and half of the molecular orbitals are anti bonding molecular orbitals.
- Anti bonding molecular orbitals have energy higher than participating atomic orbitals.
- Bonding molecular orbitals have energy lower than participating atomic orbitals.
- Molecular orbitals are that region in the molecule where electrons are most likely to found.
So, among given, option c which is 'atomic orbitals combine to form one molecular orbital' is incorrect.
Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
Answer:
C3 H6 Cl 3
Explanation:
C -24.2%
H - 4.0%
Cl - (100-24.2 - 4.0)=73.8 %
We can take 100g of the substance, then we have
C -24.2 g
H - 4.0 g
Cl - 73.8 g
Find the moles of these elements
C -24.2 g/12.0 g/mol =2.0 mol
H - 4.0 g/1.0 g/mol = 4. 0 mol
Cl - 73.8 g/ 35.5 g/mol = 2.1 mol
Ratio of these elements gives simplest formula of the substance
C : H : Cl = 2 : 4 : 2 = 1 : 2 : 1
CH2Cl
Molar mass (CH2Cl) = 1*12.0 +2*1.0 + 1*35.5 = 49.5 g/mol
Real molar mass = 150 g/mol
real molar mass/ Molar mass (CH2Cl) = 150 /49.5=3
So, Real formula should be C3 H6 Cl 3.
