The choices for this problem are bismuth, Bi; platinum, Pt; selenium, Se; calcium, Ca and copper, Cu. I think the correct answer would be selenium. The melting point of bismuth is at a temperature of 544.4 Kelvin. At a temperature of 525 K, it would exist as solid. Platinum melts at 2041.1 K. At 525 K, platinum would be in solid form. Selenium has a melting point at 494 K so that at a temperature of 525 K, it would exist in its liquid state. Calcium has a melting point of 1112 K so it would exist as solid at 525 K. Copper has a melting point at 1358 K, so it would still exist as solid at a temperature of 525 K. Therefore, the answer would only be selenium.
<span>Boron has a lot of different isotopes, most of which having a very short half life (ranging from 770 milliseconds for Boron-8 down to 150 yoctoseconds for boron-7). But the two isotopes Boron-10 and Boron-11 are stable with about 80.1% of the naturally occurring boron being boron-11 and the remaining 19.9% being boron-10. The weighted average weight of those 2 isotopes has the value of 10.81.
The reason they use the average mass of an element for it's atomic weight is because elements in nature are rarely single isotopes. The weighted average allows us to easily compare relative number of atoms of one element against relative numbers of atoms of another element assuming that the experimenters are getting isotope ratios close to their natural ratios.</span>
Answer:
Theoretical yield of C6H10 = 3.2 g.
Explanation:
Defining Theoretical yield as the quantity of product obtained from the complete conversion of the limiting reactant in a chemical reaction. It can be expressed as grams or moles.
Equation of the reaction
C6H11OH --> C6H10 + H2O
Moles of C6H11OH:
Molar mass of C6H110H = (12*6) + (1*12) + 16
= 100 g/mol
Mass of C6H10 = 3.8 g
number of moles = mass/molar mass
=3.8/100
= 0.038 mol.
Using stoichoimetry, 1 moles of C6H110H was dehydrated to form 1 mole of C6H10 and 1 mole of water.
Therefore, 0.038 moles of C6H10 was produced.
Mass of C6H10 = molar mass * number of moles
Molar mass of C6H10 = (12*6) + (1*10)
= 82 g/mol.
Mass = 82 * 0.038
= 3.116 g of C6H10.
Theoretical yield of C6H10 = 3.2 g
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.
