When a solvent has as much of the dilute dissolved in it as possible, then it is saturated.
If you were to heat the water, its capacity would increase and would then be super-saturated because it has more dissolved in it than possible as room temp.
Since there is no heating being done, the water is just saturated.
Hope that helps!
Answer:
1. B = 1.13M
2. A. 8.46%
3. D = 0.0199
Explanation:
1. Molarity of a a solution = number of moles of solute/ volume of solution in L
Number of moles of solute = mass of solute/molar mass of solute
Molar mass of NaC₂H₃O₂ = 82 g/mol, mass of NaC₂H₃O₂ = 245 g
Number of moles of NaC₂H₃O₂ = 245 g / 82 g/mol = 2.988 moles
Molarity of solution = 2.988 mols/ 2.65 L = 1.13 M
2. Percentage by mass of a substance = mass of substance /mass of solution × 100%
Mass of 2.65 L of water = density × volume
Density of water = 1 Kg/L = 1000 g/L; volume of waterb= 2.65 L
Mass of water = 1000 g/L × 2.65 L = 2650 g
Mass of solution = mass of water + mass of solute = 245 + 2650 =2895 g
Percentage by mass of NaC₂H₃O₂ = 245/2895 × 100% = 8.46%
3. Mole fraction of NaC₂H₃O₂ = moles of NaC₂H₃O₂/moles of solution
Moles of water = mass /molar mass
Mass of water = 2650 g; molar mass of water = 18 g/mol
Moles of water = 2650 g / 18 g/mol = 147.222 moles
Moles of solution = moles of solute + moles of water = 147.222 + 2.988 = 150.21
Moles of NaC₂H₃O₂ =2.988 moles
Moles fraction of NaC₂H₃O₂ =2.988/150.21 = 0.0199
Energy is distributed not just in translational KE, but also in rotation, vibration and also distributed in electronic energy levels (if input great enough, bond breaks).
All four forms of energy are quantised and the quanta ‘gap’ differences increases from trans. KE ==> electronic.
Entropy (S) and energy distribution: The energy is distributed amongst the energy levels in the particles to maximise their entropy.
Entropy is a measure of both the way the particles are arranged AND the ways the quanta of energy can be arranged.
We can apply ΔSθsys/surr/tot ideas to chemical changes to test feasibility of a reaction:
ΔSθtot = ΔSθsys + ΔSθsurr
ΔSθtot must be >=0 for a chemical change to be feasible.
For example: CaCO3(s) ==> CaO(s) + CO2(g)
ΔSθsys = ΣSθproducts – ΣSθreactants
ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s)
ΔSθsurr is –ΔHθ/T(K) and ΔH is very endothermic (very +ve),
Now ΔSθsys is approximately constant with temperature and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall.
But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800oCΔSθtot becomes plus overall (and ΔGθ becomes negative), so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.
CaCO3(s) ==> CaO(s) + CO2(g) ΔHθ = +179 kJ mol–1 (very endothermic)
This important industrial reaction for converting limestone (calcium carbonate) to lime (calcium oxide) has to be performed at high temperatures in a specially designed limekiln – which these days, basically consists of a huge rotating angled ceramic lined steel tube in which a mixture of limestone plus coal/coke/oil/gas? is fed in at one end and lime collected at the lower end. The mixture is ignited and excess air blasted through to burn the coal/coke and maintain a high operating temperature.
ΔSθsys = ΣSθproducts – ΣSθreactants
ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) = (40.0) + (214.0) – (92.9) = +161.0 J mol–1 K–1
ΔSθsurr is –ΔHθ/T = –(179000/T)
ΔSθtot = ΔSθsys + ΔSθsurr
ΔSθtot = (+161) + (–179000/T) = 161 – 179000/T
If we then substitute various values of T (in Kelvin) you can calculate when the reaction becomes feasible.
For T = 298K (room temperature)
ΔSθtot = 161 – 179000/298 = –439.7 J mol–1 K–1, no good, negative entropy change
For T = 500K (fairly high temperature for an industrial process)
ΔSθtot = 161 – 179000/500 = –197.0, still no good
For T = 1200K (limekiln temperature)
ΔSθtot = 161 – 179000/1200 = +11.8 J mol–1 K–1, definitely feasible, overall positive entropy change
Now assuming ΔSθsys is approximately constant with temperature change and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall. But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800–900oC ΔSθtot becomes plus overall, so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.
You can approach the problem in another more efficient way by solving the total entropy expression for T at the point when the total entropy change is zero. At this point calcium carbonate, calcium oxide and carbon dioxide are at equilibrium.
ΔSθtot–equilib = 0 = 161 – 179000/T, 179000/T = 161, T = 179000/161 = 1112 K
This means that 1112 K is the minimum temperature to get an economic yield. Well at first sight anyway. In fact because the carbon dioxide is swept away in the flue gases so an equilibrium is never truly attained so limestone continues to decompose even at lower temperatures.
Answer: 1) Temperature can change the solubility of a solute.
Explanation:
The chart is missing so there is no way to tell what does the graph show.
Yet, I can help you because I can explain the status of each statement of the choices. As you will see there is only one possibility..
<span>1) Temperature can change the solubility of a solute.
Yes, temperature definetly can, and mostly do, modify the solubility of a solute.
You can search any chart of solubility and will find that.
I can give you two examples:
a) Sodium chloride: dissolve some spoons of salt in a cold water until you can not dissolve more. Then, heat the water, you will find that more salt will get dissolved, proving that the temperature of the solution increases the solubility of sodium chloride.
b) Carbon dioxide gas: the soft drinks have CO₂ molecules dissolved in it.
The higher the temperature of the soft drink the less the amount of CO₂(g) that can be dissolved. That is why the soda bottling plants cool the beverage before adding the CO₂(g).
2) </span><span>Temperature has no affect on the solubility of a solute.
Since this is the opposite to the first statement and the first is true, this is false.
3) Salt has a greater solubility than sugar.
False.
This is an empirical result, which you cannot predict theoretically. So you need to see at the data either in a table or in a chart. Else you can test it at home. After the empirical data are shown it results that more grams of sugar can be dissolved in water compared to salt.
That is something you ca see in a chart or you can prove by yourself.
4) Nitrite salt has a greater solubility than sugar.
</span>
False.
Looking at some data you can find that sodium nitrite solutiliby is aroun 70 - 100 g/10 g while sugar (sucrose) solutiblity is around 180 - 235 g/ 100 g.
This must be your work, otherwise it's against the Brainly code of conduct. If you need help finding a newspaper, go to BBC. They have plenty of relevant articles.
