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Luden [163]
3 years ago
11

When the vessel and its contents are warmed to 100 °C, Q decomposes into its constituent elements. What is the total pressure, a

nd what are the partial pressures of xenon and oxygen in the container?

Engineering
1 answer:
Aleksandr [31]3 years ago
4 0

Answer:

The question is incomplete, but here is a similar question with complete details ; Gaseous compound Q contains only xenon and oxygen . When 0.100 g of Q is placed in a 50.0 mL steel vessel at 0 °C, the pressure is 0.229 atm. (a) What is the molar mass of Q (b) When the vessel and its contents are warmed to 100 °C, Q decomposes into its constituent elements . What is the total pressure , and what are the partial pressures of Xenon and Oxygen

Explanation:

The step by step calculations is as shown in the attached file.

It should be noted that what was applied is

  1. The Ideal gas equation PV = nRT
  2. Daltons law of partial pressure which states that in a mixture of gases, the total pressure exerted is equal to the sum of the individual partial pressures of the gases at constant temperature.
  3. It should be noted that the total pressure of the gases can be gotten by applying pressures law at constant volume
  4. P1/P2 = T1/T2
  5. It should also be noted that Partial pressure = Total pressure x Mole fraction

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Calculate the osmotic pressure of seawater containing 3.5 wt % NaCl at 25 °C . If reverse osmosis is applied to treat seawater,
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Answer:

Highest osmotic pressure that membrane may experience is

' =58.638 atm

Explanation:

Suppose sea-water taken is M= 1 kg

Density of water = 1000 kg/m3

Therefore Volume of water= Mass,M/Density of water

V= 1 kg/(1000 kg/m3)

V= 10-3 m3= 1 Litre

Since mass of Nacl is 3.5 wt%,Therefore in 1 kg of water

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Thus its Number of moles of Nacl= m/M.W

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= 0.5989 mol

ans since volume of solution is 1 L thus concentration of NaCl is ,C= number of moles/Volume of solution in Litres

C= 0.5989mol/ 1L

=0.5989 M

Since 1 mol NaCL disssociates to form 2 moles of ions of Na+ andCl- Thus van't hoff factor i=2

And osmotic pressure  = iCRT ------------------------------(1)( Where R= 0.0821 L.atm/mol.K and T= 25oC= 298.15 K)

Putting in equation 1 ,we get  = 2*(0.5989 mol/L)*(0.0821 L.atm/mol.K)*298.15 K

=29.319 atm

Now as the water gets filtered out of the membrane,the water's volume decreases and concentration C of NacL increases, thus osmotic pressure also increases.Thus, at 50% water been already filtered out, the osmotic pressure at the membrane will be maximum

Thus Volume of water left after 50% is filtered out as fresh water= 0.5 L (assuming no salt passes through semi permeable membrane)

Thus New concentration of NaCl C'= 2*C

C'=2*0.5989 M

=1.1978 M

and Since Osmotic pressure is directly proportional to concentration, Thus As concentration C doubles to C', Osmotic Pressure  ' also doubles from  ,

Thus,Highest osmotic pressure that membrane may experience is,  '=2*  

=2*29.319 atm

' =58.638 atm

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