How many liters of water must be added to 5.40 g of sodium nitrate to create a solution that has a concentration of 3.81 g/l?
grams of solute
use 9
liter of solution
08
0.706 l
1.42 l
1.59 l
20.6 l
0

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

Hi + naoh -> h2o + nal

The answer is c.  h+  and oh−  react to form water (h2o). i just took the test and i got it right. hope this !

B.1 - hydrogren-1 can only mean there is 1 neutron. it's obvious there is 1 proton or else it wouldn't be hydrogen, so they wouldn't put a number after it unless they were indicating the number of neutrons