Answer:
Potassium iodide increases the decomposition rate of hydrogen peroxide.
Explanation:
Potassium iodide increases the decomposition rate of hydrogen peroxide because potassium iodide act as a catalyst. A catalyst speed up the process of chemical reaction without reacting with the molecules present in reaction. If the potassium iodide is not present as a catalyst for the decomposition of hydrogen peroxide then the decomposition of hydrogen peroxide takes too much time because the catalyst is absent that speed up the reaction.
The correct answer is Cl₂ <CCl₄ < PCl₃ < CaCl₂ < CsCl (most polar)
Cl₂ is nonpolar or least polar as there is no differentiation in electronegativity of the two atoms of chlorine producing amongst them. CCl₄ exhibits four C-Cl bonds. The C-atom is electropositive and Cl is electronegative. Therefore, there are four dipoles in CCl₄. Though, these dipoles cancel each other because of the tetrahedral geometry of the molecule. Therefore, CCl₄ is less polar in comparison to others, however, more polar in comparison to Cl₂.
PCl₃ exhibits three dipoles because of electronegative chlorine atoms, and electropositive phosphorus atom. Though, the configuration of the molecule is trigonal pyramidal, producing it more polar in comparison to CCl₄. The CaCl₂ is ionic compound, therefore, polar in comparison to the covalent bond compounds.
CsCl is ionic compound, however, more polar in comparison to CaCl₂, as there is the higher difference in electronegativity of Cs and Cl compared with that between the Ca and Cl.
Answer:
K3PO4
Explanation:
Recall that colligative properties depends on the number of particles present. The greater the number of particles present, the greater the degree of colligative properties of the solution. Let us look at each option individually;
SrCr2O7-------> Sr^2+ + Cr2O7^2- ( 2 particles)
C4H11N (not ionic in nature hence it can not dissociate into ions)
K3PO4-------> 3K^+ + PO4^3- (4 particles)
Rb2CO3-------> 2Rb^+ + CO3^2- (3 particles)
Hence K3PO4 has the greatest number of particles and will display the greatest colligative effect.
