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luda_lava [24]
3 years ago
9

Which of the following atoms could have an expanded octet when it is the central atom in a covalent compound?a) B

Chemistry
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

only chlorine can expand its octet.

Explanation:

An atom can expand its octet is it has empty d orbital

the electronic configuration of given elements will be:

B : 1s2 2s2 2p1 [Valence shell n =2 no d orbital]

O :1s2 2s2 2p4 [Valence shell n =2 no d orbital]

F : 1s2 2s2 2p5 [Valence shell n =2 no d orbital]

Cl :1s2 2s2 2p6 3s2 3p5 3d0 [Valence shell n =2 no d orbital]

Out of given elements only chlorine has empty d orbitals in its valence shell

Thus only chlorine can expand its octet.

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The mass of 254 mL of water is 254 g. Since the density of water is 1g/mL, we can simply multiply the density 1g/mL by 254 mL of water and get 254 g as our answer.  Since mL is in the numerator and denominator, mL cancels out and we are left with g only. 
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Strontium-90 is a radioisotope that will decrease in mass by one-half every 29 years. How many years will it take for a 10.0-gra
Lelechka [254]

Answer: It will take 29 years for a 10.0-gram sample of strontium-90 to decay to 5.00 grams

Explanation:

Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life is represented by t_{\frac{1}{2}

t_{\frac{1}{2}=\frac{0.693}{\lambda}

\lambda = rate constant

Given : Strontium-90 decreases in mass  by one-half every 29 years , that is half life of Strontium-90  is 29 years.

As half life is independent of initial concentration, it will take 29 years for a 10.0-gram sample of strontium-90 to decay to 5.00 grams as the amount gets half.

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To what volume in millimeters must 50.0 mL of 18.0 M H2SO4 be diluted to obtain 4.35 M H2SO4?
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We know that to relate solutions of with the factors of molarity and volume, we can use the equation: M_{1}  V_{1} = M_{2}  V_{2}

**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.

So now we can assign values to these variables. Let us say that the 18 M H_{2}  SO_{4} is the left side of the equation. Then we have:

(18 M)(0.050 L)=(4.35M) V_{2}

We can then solve for V_{2}:

V_{2}= \frac{(18M)(0.05L)}{4.35M} and V_{2} =0.21 L or 210 mL

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At what temperature is the following reaction feasible: HCl(g) + NH3(g) -> NH4Cl(s)?
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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.
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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.

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