Answer is: adding NaCl will lower the freezing point of a solution.
A solution (in this example solution of sodium chloride) freezes at a lower temperature than does the pure solvent (deionized water).
The higher the solute concentration (sodium chloride), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
Dissociation of sodium chloride in water: NaCl(aq) → Na⁺(aq) + Cl⁻(aq).
Answer:I think it’s A.Positive. Not sure though.
Explanation:
Describe the process by which Ag+ ions are precipitated out of solution. 4. In your testing, several precipitates are formed, and then dissolved as complexes.
The type of radiation which is identical to a high energy electron is known as a(n) beta.
Answer: A) or the first option.
In an ionic compound the atoms are linked via ionic bonds. These are formed by the transfer of electrons from one atom to the other. The atom that loses electrons gains a positive charge whereas the atom that accepts electrons gains a negative. This happens in accordance with the octet rule wherein each atom is surrounded by 8 electrons
In the given example:
The valence electron configuration of Iodine (I) = 5s²5p⁵
It needs only one electron to complete its octet.
In the given options:
K = 4s¹
C = 2s²2p²
Cl = 3s²3p⁵
P = 3s²3p³
Thus K can donate its valence electron to Iodine. As a result K, will gain a stable noble gas configuration of argon while iodine would gain an octet. This would also balance the charges as K⁺I⁻ creating a neutral molecule.
Ans: Potassium (K)
