Answer:
The answer is (A)
Explanation:
When the weather changes, nature also changes because most plants rely on photosynthesis and if they don't get as much light then they can't support as much as they used causing them to shut down parts of the plant.
The question is missing the graphics required to answer which I have attached as an image.
There are four different representations of the orientation of water molecules around chloride anion. Let's first analyze the water molecule.
We have H-O-H as the structure of water. The oxygen atom is more electronegative than the hydrogen atoms, which results in a partial positive charge on the hydrogen atoms and a partial negative charge on the oxygen atom.
The chloride anion is a negative charge. Therefore, the water molecules should orient themselves with the hydrogen atoms facing the chlorine atom as the partial positive charge on the hydrogen atoms will be attracted to the negative charge of the chlorine atom.
The correct representation is shown in graph 3 which shows all hydrogen atoms facing the chlorine anion.
Answer:
a. The atom will go from a two-dimensional configuration to a three dimensional configuration.
d. The bond angle will increase.
f. The number of unhybridized p orbitals will decrease.
Explanation:
Sp2 is the atomic bond in which orbitals mixes with only two orbitals. These orbitals form three sp2. When two carbon atoms are overlapped they form sigma bond by overlapping of sp2 bonds. Sp3 bond is created when there is one lone molecule available for combination. When the bonding is updated from sp2 to sp3 then unhybridized orbitals will decrease causing the bond angle to increase.
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K