Answer: at relatively low temperatures.
Explanation:
According to Gibbs equation;
= Gibb's free energy change
= enthalpy change
T = temperature
= entropy change
A reaction is spontaneous when = Gibb's free energy change is negative.
Thus 
Thus the reaction is spontaneous or is negative only when 
Thus the reaction is spontaneous at relatively low temperatures
Answer:
SO₂
Explanation:
The dipole-dipole force is not only determined by the electron density around each atoms in the molecule (dependent of electronegativity difference), but also how the atoms in the molecules are arranged. In general, the more symmetric a molecule is, the less dipole force it exerts as each dipole moments cancels each other out.
Now let's examine each answer
- b and c, N₂ and H₂ are composed of same type of atoms, therefore, no dipole moment occurs, and no dipole-dipole forces are exerted
- a and e, BCl₃ and CBr₄ are composed of atoms with different electronegativity, but are symmetric. With BCl₃ having trigonal planar structure and CBr₄ has tetragonal structure, each B-Cl and C- Br bond cancels out each other dipole moment, and thus, no dipole moments were generated.
- d, SO₂ structure is not linear, but is a little bent, this allows net dipole moment to occurs in this molecule as dipole moment from each S=O bond do not cancels each other out
Answer: The molecule identified with the letter C. HFO.
Explanation:
1) In covalent bonds, the dipoles are identified with an arrow pointing from the least electronegative element toward the most electronegative element.
2) The molecule HFO is formed by one covalent bond between H and O, and other covalent bond between F and O.
3) O is more electronegative than H, so in the H-O bond the dipole is identified with the arrow from the H atom toward the O atom. That is correctly indicated in the drawing C, in this way:
-|----→
H - O
4) F is more electronegative than O, so in the O-F bond the dipole is identified with an arrow from the O atom toward the F atom. That is correctly indicated in this way:
-|---→
O - F.
Step (1):
Generation of electrophile: by the action of Lewis acid FeCl₃ on Cl₂ to serve as a source of Cl⁺ (Electrophile)
Step (2):
Addition of electrophile to form carbocation:
addition of electrophile to form C-Cl bond and form carbocation which is stabilized by resonance.
Step (3):
Loss of proton to re-form the aromatic ring by the action of FeCl₄⁻ which removes proton from carbon containing Cl and forming the aromatic ring again
Answer: the solution 1.0 M Ca Br2.
This is becasue the freezing point depression is a colligative property.
A collegative property is one that depends on the number of particles in solution.
Each mole of KBr will produce two moles, one mole of potassium ion and one of bromine ion
Each mole of CaBr2 will produce three moles, one of calcion ion and two of bromine ion.
Then, the depression of the frezzing point will be greater in the case of the CaBr2, leading to a lower freesing point of the 1.0 M soluction.
