Answer:
It is made up of atoms that are held together by covalent bonds.
Explanation:
The correct question is
Give the coordination number, the charge of the central metal ion, and select the correct name in each coordination compound:
Na3[CoCl6]
[Ni(CO)4]
[Ni(NH3)3(H2O)2] (NO3)2
Answer:
See explanation for details
Explanation:
Na3[CoCl6]
Name: Sodium hexacholorocobalt III
Charge on the complex: -3
Central metal ion name : cobalt III
coordination number: 6
[Ni(CO)4]
Name: tetracarbonyl nickel (0)
Charge on the complex: 0
Central metal ion name : nickel (0)
coordination number: 4
[Ni(NH3)3(H2O)2] (NO3)2
Name: Diaquatriaminenickel II nitrate
Charge on the complex: +2
Central metal ion name : nickel (+2)
coordination number: 5
The triprotic acid like H₃PO₄ contains three protons H⁺ so it ionized through three steps, the acidity strength of first proton higher than second proton higher than third one so Ka1 > Ka2 > Ka3 But
For pKa which equals to -log Ka ... the higher the value of Ka, the smaller the value of pKa. so the correct answer will be:
pKa1 < pKa2 < pKa3
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.
43.62m
Explanation:
Given parameters:
Energy = 8550J
Mass of bell = 20kg
Unknown:
Height of the bell = ?
Solution:
As this state, the bell possesses potential energy. Potential energy is the energy at rest in a body. It is due to the position of the body.
Potential energy = m g h
m is the mass of the body
g is the acceleration due to gravity of the body
h is the height of the body
The unknown is h and we should solve for it.
g = 9.8m/s²
Input the variables;
8550 = 20 x 9.8 x h
h = 43.62m
learn more:
Potential energy brainly.com/question/10770261
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