<em>Answer:</em>
- The concentration of new solution will be 1×10∧-7 M.
<em>Solution:</em>
<em>Data Given </em>
given mass of fluoxymesterone =16.8mg = 0.0168 g
molar mass of fluoxymesterone = 336g/mol
vol. of fluoxymesterone = 500.0 ml = 0.500 L
Stock Molarity of fluoxymesterone = (0.0168/336)÷0.500 = 1×10∧-4 M
So applying dilution formula
Stock Solution : New Solution
M1.V1 = M2.V2
( 1×10∧-4 M) × (1×10∧-6 L) = M2 × 0.001 L
[( 1×10∧-4) × (1×10∧-6)]÷[0.001] = M2
1 × 10∧-7 = M2
<em>Result:</em>
- The concentration of new solution M2 will be 1 × 10∧-7
Answer:
you can do anything if you put your mind to it. Also, school is not everything in life there are amazinng things in life that you need school for. you got this.
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: matter and its interaction with energy. Matter has mass and occupies space. A Substance is matter that cannot be separated into other kinds of matter by physical means.
Explanation:
Answer: hello your question lacks some data attached below is the missing data
answer :
a) 3-methyl heptane
b) 2-methyl pentane
c) 2-methyl heptane
d) 2-methyl hexane
e) 3-methyl hexane
Explanation:
we will select the longest carbon chain as the branched alkane and name it
a) 3-methyl heptane ( first diagram )
b) 2-methyl pentane ( second diagram )
c) 2-methyl heptane ( third diagram )
d) 2-methyl hexane ( fourth diagram )
e) 3-methyl hexane ( fifth diagram )
<em>Note : sixth diagram = first diagram </em>
