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Rufina [12.5K]
3 years ago
5

.) Neon and HF have approximately the same molecular masses. a.)Explain why the boiling points of Neon and HF differ b.)Compare

the change in the boiling points of Ne, Ar, Kr, and Xe with the change of the boiling points of HF, HCl, HBr, and HI, and explain the difference between the changes with the increasing atomic or molecular mass.
Chemistry
1 answer:
lisov135 [29]3 years ago
4 0

Answer:

See explanation

Explanation:

a) The magnitude of intermolecular forces in compounds affects the boiling points of the compound. Neon has London dispersion forces as the only intermolecular forces operating in the substance while HF has dipole dipole interaction and strong hydrogen bonds operating in the molecule hence HF exhibits a much higher boiling point than Ne though they have similar molecular masses.

b) The boiling points of the halogen halides are much higher than that of the noble gases because the halogen halides have much higher molecular masses and stronger intermolecular forces between molecules compared to the noble gases.

Also, the change in boiling point of the hydrogen halides is much more marked(decreases rapidly)  due to decrease in the magnitude of hydrogen bonding from HF to HI. The boiling point of the noble gases increases rapidly down the group as the molecular mass of the gases increases.

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1. A sample of oxygen is collected over water at 22 ° C and 762 torr. What is the partial pressure of the dry oxygen? The vapor
Rom4ik [11]

Answer: The partial pressure of the dry oxygen is 742 torr

Explanation:

Dalton's Law of Partial Pressure states that the total pressure exerted by a mixture of gases is the sum of partial pressure of each individual gas present. Thus P(total)=P_1+P_2 .........

Given; Total pressure = 762 torr

partial pressure of water = 19.8 torr

partial pressure of dry oxygen = ? torr

Total pressure  = partial pressure of water + partial pressure of dry oxygen

762 torr = 19.8 torr = partial pressure of dry oxygen

partial pressure of dry oxygen = 742 torr

The partial pressure of the dry oxygen is 742 torr

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4 years ago
At what temperature is the following reaction feasible: HCl(g) + NH3(g) -> NH4Cl(s)?
Nutka1998 [239]
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.

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Answer:

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Explanation:

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V = 300 mL = 0.3 L

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M=\frac{m}{V}=\frac{0.6}{0.3}=2.00 M

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