Answer:
The molecule has a bent geometry
Explanation:
Let us look again at the principles of VSEPR theory. The shape of a molecule depends on the number of electron pairs that surround the valence shell of the central atom in the molecule.
Lone pairs distort the molecular geometry away from what is expected on the basis of VSEPR theory.
The molecule described in the question has the form AEX2. Two substituents and one lone pair form three electron domains around the central atom. The expected geometry is trigonal planar but the observed molecular geometry is bent because of the lone pairs present.
Answer:
Explanation:
At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).
Hence, the <em>complete combustion reaction</em> that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.
Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).
A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.
<u>a. C₂H₄:</u>
- C₂H₄ (g) + 3O₂ (g) → 2CO₂(g) + 2H₂O (g)
Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.
The following analysis just shows that the other options are not right.
<u>b. C₂H₂:</u>
- 2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g) + 2H₂O (g)
The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.
<u>с. С₃Н₈</u>
- C₃H₈ (g) + 5O₂ (g) → 3CO₂(g) + 4H₂O (g)
The mole ratio is 1 mol C₃H₈ : 5 mol O₂
<u>d. C₂H₆</u>
- 2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g) + 6H₂O (g)
The mole ratio is 2 mol C₂H₆ : 7 mol O₂
Answer:
0.46 V
Explanation:
The emf for the cell is given by:
Eº cell = Eº oxidation + Eº reduction
From the given balanced chemical equation, we can deduce that Fe²⁺ has been oxidized to Fe³⁺, and O reduced from 0 to negative 2, according to the half cell reactions:
4Fe²⁺ ⇒ Fe³⁺ + 4e⁻ oxidation
O₂ + 4H⁺ + 4 e⁻ ⇒ 2 H₂O reduction
From reference tables for the standard reduction potential, we get
Eº red Fe³⁺ / Fe²⁺ Eºred = 0.77 V
Eº red O₂ / H₂O Eºred = 1.23 V
Now all we need to do is change the sign of Eº reduction for the species being oxidized ( Fe²⁺ ) and add it to Eº reduction O₂:
Eº cell = Eº oxidation + Eº reduction = - (0.77 V ) + 1.23 V = 0.46 V
Answer:
butyne
Explanation:
alkane, alkene, and alkyne are all examples of hydrocarbons.
butyne = alkyne
