Answer:
<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>
Explanation:
This question is about solubility.
Regarding solubility, the solutions may be classified as:
- Unsaturated: the concentration is below the maximum concentration permited at the given temperature.
- Saturated: the concentration is the maximum permitted at the given temperature, under normal conditions.
- Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.
Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.
- In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.
- In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g of water.
- Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ
115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O ⇒ x = 57.5 g NaNO₃
- <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>
<em />
Physical. chemical only applies when it undergoes a reaction
Answer:
less gravity, closest to the sun
In order to determine the concentration of ammonium ions in
the solution prepared by mixing solutions of ammonium sulfate, (NH4)2SO4, and ammonium
nitrate, first calculate the amount of ammonium ions for each solution.<span>
<span>For ammonium sulfate sol'n: 0.360 L x 0.250 mol(NH4)2SO4/ L x 2 mol NH4+ /1 mol(NH4)2SO4 =
0.18 mol NH4+
<span>For ammonium nitrate sol'n: 0.675 x 1.2 mol NH4NO3/L x 1 mol NH4+ /1 molNH4NO3
= 0.81 mol NH4+
Thus, the amount of NH4+ ions is (0.18 + 0.81) mol or 0.99
mol NH4+. To get the concentration, multiply this to the volume of solution
which is assumed to be additive, such that:</span></span></span>
M NH4+ in sol’n = 0.99 mol NH4+/1.035 L = 0.9565 mol NH4+/ L
sol’n
The reaction yields the product compound 2,4,5-tribromophenol.
<h3>Bromination of monohydroxy benzene</h3>
Bromination is an organic reaction in which aqueous bromine is made to react with another compound so as to introduce bromine into the molecule.
When we carry out the bromination of monohydroxy benzene, we obtain the compound 2,4,5-tribromophenol.
Learn more about benzene: brainly.com/question/14525517
