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WARRIOR [948]
2 years ago
9

When is the small size of gas particles taken into account?

Chemistry
1 answer:
jeyben [28]2 years ago
3 0

Answer:

At high pressures and low temperatures.  

Explanation:

That's when the volume of the gas is quite small.

The volume of the gas particles can then be a significant proportion of the total volume.

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There are two opposing processes that occur in a solution in contact with undissolved solute. These are dissolving and ________.
Alex777 [14]

Answer:

c. crystallization

Explanation:

The opposing process that occur in a solution in contact with undissolved solute are dissolution and crystallization.

In the dissolution process the solid substance coverts into liquid state and mixes with solution. Whereas in Crystallization the the chemical is converted from the liquid solution to solid crystal state.

3 0
3 years ago
If the equation: FeCl3+O2-->Fe2O3+Cl2, how many moles of chlorine gas can be produced if 4 moles of FeCl3 react with 4 moles
BlackZzzverrR [31]

Answer:

6 moles of Cl2

Explanation:

First, the equation has to be balanced, which makes it 4 FeCl3 + 3 O2 --> 2 Fe2O3 + 6 Cl2

Using this information, we can see that one mole of O2 will not be present in the reaction. Since four moles of FeCl3 are needed to react in the equation, which would produce six moles of Cl2, and only four moles of FeCl3 are present, six moles of Cl2 would be produced.

4 0
2 years ago
If the air temperature is the same as the temperature of your skin (about 30◦C), your body cannot get rid of heat by transferrin
Elenna [48]

Answer:

Explanation:

a )

energy produced per second = 500 J

Heat produced = 500 x .8 = 400 J per second.

If m be the mass of water evaporated per unit hour

m x latent heat = 400 x 60 x 60

= m x 2.42 x 10⁶ = 1.44 x 10⁶

m = .595 kg per hour

b )

volume of water = 595 mL

bottles = 595 / 750

.8 or 4/5 of bottle. per hour.

8 0
3 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.

8 0
2 years ago
Read 2 more answers
Niobium-91 has a half-life of 680 years. After 2,040 years, how much niobium-91 will remain from a 300.0-g sample? 3 g 18.75 g 3
Vadim26 [7]
Amount of Niobium-91 initially
= 300/91 =3.2967mol
2040 years = 3 ×680 = 3 half-lives
therefore, amount left = 0.4121mol
mass of Niobium-91 remaining = 0.4121 ×91 =37.5g
7 0
3 years ago
Read 2 more answers
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