Answer: I don't think you can delete questions tho sorry
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.
<span>copper(II) chloride octahydrate</span>
Look it up on the periodic table i attached below
im too lazy to type that all out im sorry but this is one of the easiest things you'll do in chemistry and if you tried you could actually finish all 5 of these in less than a minute instead of taking like 10 min to type 10 problems out separately and making 4 diff posts