1.one substance must dissolve another : the solvent breaks the the solute causing it to dissociate into the solvent it can be both solids or gases that dissolve in the solute
2.the melting point will be higher than 100 degrees as the salt causes the solution to gain a higher melting point
3.either released or absorbed : energy is required to dissolve the solute but the solute may also release energy
4.the salt dissolved most quickly in the warm water : as there is more energy for the solute to dissolve
hope that helps
Answer:

Explanation:
The given reactions are:
PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq) 
AgCl(aq)⇌Ag+(aq)+Cl−(aq) 
Required reaction is:
PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(aq)

Answer:
<u>Asexual reproduction</u> = An exact copy of the reproducer
<u>Sexual reproduction</u> = Genetic variation into the organism if a random mutation in the organism's DNA is transmitted to offspring.
Explanation:
Example of Asexual Reproduction diversity - A red apple has apple seeds. You plant the red apple seeds, and it grows up to be no different than the apple before it.
Sexual reproduction Example of diversity -Two parents are of different ethnicities. The female gets pregnant, but because of the different genetics from both of the parents, the child will inherit their genetics + some random mutation in the DNA.
This impacts the diversity of the offspring because it can be an exact copy of the reproducer (Asexual), or significantly different with some similarities if the mode of reproduction was sexual.
I really hope this helps you! Please tell me if it did or not. Good luck with your assignment/exam/quiz!
(OMG TYSM FOR BRAINLIEST!! :D)
First, we need to get the value of Ka:
when Ka = Kw / Kb
we have Kb = 1.8 x 10^-5
and Kw = 3.99 x 10^-16 so, by substitution:
Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11
by using the ICE table :
NH4+ + H2O →NH3 + H+
intial 0.013 0 0
change -X +X +X
Equ (0.013-X) X X
when Ka = [NH3][H+] / [NH4+]
by substitution:
2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X
∴X = 5.35 x 10^-7
∴[H+] = X = 5.35 x 10^-7
∴PH = - ㏒[H+]
= -㏒(5.35 x 10^-7)
= 6.27