Answer:
This is due to more hydrogen bonding in ethylene glycol than it is in isopropyl alcohol
Explanation:
The boiling point of isopropyl alcohol is 82.4 °C it contains only a single OH group, hence intermolecular hydrogen bonding is solely responsible for it's boiling point, whereas Ethylene glycol (CH2OHCH2OH) contains 2-OH group and both intermolecular and intramolecular hydrogen bonding are responsible for the higher boiling point of ethylene glycol at 198 °C.
Not sure what you are asking. I have two possible answers though...
It could either be more negatively charged, or valence electrons.
The more away from the nucleus a electron is, the more negatively charged it is.
The electrons on the outermost electron shell is valence electrons.
Again, I don't know what you were asking, but one of these answers may be correct.
Your question isn't quite clear, but if you're wondering if a chemical is polar or non-polar, you simply draw a VSEPR sketch and draw arrows where the bonds are. Only draw arrows between atoms, NOT between an atom and a lone pair of electrons. The arrow should point to the most electronegative atom (you should be given an electronegativity scale). Afterwards, you add up the arrows as vectors, and look at the sum of the vectors. If the sum is zero (CH4 is a good example), the chemical is non-polar. If the sum is a vector, the chemical is polar (H2O, or water, is polar).
Answer:
The solutions are ordered by this way (from lowest to highest freezing point): K₃PO₄ < CaCl₂ < NaI < glucose
Option d, b, a and c
Explanation:
Colligative property: Freezing point depression
The formula is: ΔT = Kf . m . i
ΔT = Freezing T° of pure solvent - Freezing T° of solution
We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.
Option d, which is glucose is non electrolyte so the i = 1
a. NaI → Na⁺ + I⁻ i =2
b. CaCl₂ → Ca²⁺ + 2Cl⁻ i =3
c. K₃PO₄ → 3K⁺ + PO₄⁻³ i=4
Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.
