Question

Q) 1)
a)
A gas cylinder contains 60% N₂ , 15%O₂ and 25%CO₂ at 760mm Hg pressure.Calculate the partial pressure of each gas.

b)
Write The configuration of O₂,O₂⁺,O₂²⁺ molecular species. Arrange them in increasing order of
i) Bond Length
ii) Stability
iii) magnetic behaviour

c)
Consider a classroom that is roughly 5 m x 10 m x 3 m. Initially, the temperature = 20.°C and the pressure = 1.0 atm. There are 50 people in class studying physical chemistry. Each person loses energy to the room at the average rate of 150 watts. Assume that the walls, ceiling, floor, and furniture are perfectly insulated and do not absorb heat. How many minutes will the physical chemistry class last if the professor has agreed to dismiss the class when the air temperature in the room reaches body temperature, 37°C? Cp=7R/2. Loss Of air to the outside as the temperature rises may be neglected.(R=8.314J/K mol).

d)
The standard heat of formation of Fe₂O₃(s) is -824.2 kJ/mol. Calculate the Heat Change for The Reaction. 4Fe(s) + 3O₂(g) ⇒ 2Fe₂O₃(s)

e)
How many photons per second Of Infrared radiation are produced by an infrared lamp that consumes energy at the rate 100 watt (100J/s) and is 12% efficient in converting this energy to infrared radiation? Assume that the radiation has a wavelength of 1500nm.


First Right Answer Will Be Marked As Brainliest.....!!!

Answer 1

Answer:

1 According to Dalton's law of partial pressure:

Moles of Nitrogen = 2855=1.964

Moles of Oxygen = 3220=0.625

Moles of CO2=4425=0.568

Total moles = 3.157

P′=mole fraction×PM

PN2=3.1571.964×760=472.8 mm

PO2=3.1570.625×760=150.4 mm

PN2=3.1570.568×760=136.7 mm

The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length.

so, we know how to calculate the bond order….actually let me just show it

(B.O)=1/2[(no of electron in anti bonding MO)-(no.of electrons in bonding MO)]

for, O2 : BO=1/2[10–6]=2

for, O2- : BO=1/2[10–7]=1.5

for, O2+ : BO=1/2[10–5]=2.5

for, O2^–2 : BO=1/2[10–8]=1

for, O2^+2 :BO=[same as N2]=3

So, the increasing sequence bond order

O2^-2 < O2- < O2 < O2+ < O2^+2

therefore, the increasing order of bond length of these species is,

O2^+2 < O2+ < O2 < O2 < O2^-2

3 i dont know other 3

Answer 2

Answer:

1 According to Dalton's law of partial pressure:

Moles of Nitrogen = 2855=1.964

Moles of Oxygen = 3220=0.625

Moles of CO2=4425=0.568

Total moles = 3.157

P′=mole fraction×PM

PN2=3.1571.964×760=472.8 mm

PO2=3.1570.625×760=150.4 mm

PN2=3.1570.568×760=136.7 mm

The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length.

so, we know how to calculate the bond order….actually let me just show it

(B.O)=1/2[(no of electron in anti bonding MO)-(no.of electrons in bonding MO)]

for, O2 : BO=1/2[10–6]=2

for, O2- : BO=1/2[10–7]=1.5

for, O2+ : BO=1/2[10–5]=2.5

for, O2^–2 : BO=1/2[10–8]=1

for, O2^+2 :BO=[same as N2]=3

So, the increasing sequence bond order

O2^-2 < O2- < O2 < O2+ < O2^+2

therefore, the increasing order of bond length of these species is,

O2^+2 < O2+ < O2 < O2 < O2^-2

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