Answer:
1. light is made up of energy
2. light travels in a straight line
3. the speed of light is exactly 299 792 km per second
4. this is the speed when light is traveling in a vacuum and not obstructed by the atmosphere 5. traveling at the speed of light you could go around the earth seven and a half times in a second
6. light can move super fast super slow and not at all
7. we can use light to weigh Stars
8. we can use light as tweezers
9. bubbles can turn sound into light
10. lasers can make things cold
It is definitely not A. B is an effect. I would say C because D is more of a conservative answer , C is more of a liberal answer, and we currently live in a liberally swayed world. They are probably looking for C. It is not in your nature to be bad.
Answer:
Give any two application for isotopes
Isotopes of iodine are used for radiotherapy in treatment of hyperthyroidism, cancer, etc.
Uranium, Radium, Polonium isotopes are used in atomic reactors.
Cobalt isotopes are used for irradiation of food products.
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:
The value of Kc for the reaction is 3.24
Explanation:
A reversible chemical reaction, indicated by a double arrow, occurs in both directions: reagents transforming into products (
direct reaction) and products transforming back into reagents (inverse reaction)
Chemical Equilibrium is the state in which direct and indirect reactions have the same reaction rate. Then taking into account the rate constant of a direct reaction and its inverse the chemical constant Kc is defined.
Being:
aA + bB ⇔ cC + dD
where a, b, c and d are the stoichiometric coefficients, the equilibrium constant with the following equation:
![Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%20%2A%5BB%5D%5E%7Bb%7D%20%7D)
Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reagents also raised to their stoichiometric coefficients.
Then, in the reaction 3A(g) + 2B(g) ⇔ 2C(g), the constant Kc is:
![Kc=\frac{[C]^{2} }{[A]^{3} *[B]^{2} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7B2%7D%20%7D%7B%5BA%5D%5E%7B3%7D%20%2A%5BB%5D%5E%7B2%7D%20%7D)
where:
- [A]= 0.855 M
- [B]= 1.23 M
- [C]= 1.75 M
Replacing:
Solving you get:
Kc=3.24
<u><em>The value of Kc for the reaction is 3.24</em></u>
