Answer:What is the question
Explanation:
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.
The <span>molar concentration of the crystal violet solution is more concentrated than that of the sodium hydroxide solution. It is because the crystal violet solution has more solute in it compared to the sodium hydroxide.</span>
<span>First:
46.7 g of N with 53.3 g of O,
=> mass ratio O to N = 53.3 / 46.7 = 1.1413
Second
17.9 g of N and 82.0 g of O.
mass ratio of O to N = 82.0 / 17.9 = 4.5810
Third
Ratio of the mass ratio of O to N in the second compound
to the mass ratio of O to N in the first compound =
= 4.5810 / 1.1413 = 4.013 ≈ 4
Answer: 4
</span>
H2SO4 + 2RbOH -> Rb2SO4 + 2H2O
If you want an explanation, keep reading.
In the first portion, there are two hydrogen ions and four sulfate ions.
The second portion has one rubidium ions and one hydroxide ion.
On the other side of the equation, in order to keep those two rubidiums balanced, you'll need to add a two at the beginning of the second portion, but in that process you are giving a second hydroxide value.
Back to the right side, there is there is water (H2O).
On the first portion, there were two hydrogen ions. The second portion also has two hydroxides because of the value change (adding the two to the front).
So on the fourth portion, you'd have to add another two so you could balance the four hydrogen ions (H2 and 2OH) and the two oxygen ions (2OH).
I hope this was easy to understand.