I believe the answer is Canada!
Answer:
One mole
Explanation:
It tells u the concentration
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:
Volume of the sulfuric acid (25cm³), same mass of each metal (1g)
Explanation:
In an experiment, the CONTROL VARIABLE also known as constant is the variable that is kept unchanged for all groups in an experiment. This is done in order not to influence the outcome of the experiment.
In this case, students are trying to investigate the reactivity of four different metals. They added 1 g of each metal to 25cm³ of sulfuric acid and recorded the temperature change. Based on the explanation of control variable above, the VOLUME OF SULFURIC ACID (25cm³) and the MASS OF EACH METAL (1g) are the CONTROL VARIABLES because they are the same or unchanged in this experiment.
