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3 years ago

The mole fraction of oxygen in dry air near sea level is 0.20948. The concentration of

2 answers:

6 0

N sub air = ( P ) ( V ) / ( R ) ( T )

n sub air = ( 739 / 760 ) ( 1.000 ) / ( 0.08205 ) ( 28.5 + 273.2 ) 

n sub air = 0.03928 mol per L 

R = universal gas constant = 0.08205 atm - L per mol - deg K 

Find moles O2 per liter : 
--------------------------------------... 

n sub O2 = ( y sub O2 ) ( n sub air ) 

n sub O2 = ( 0.20948 ) ( 0.03928 ) 

n sub O2 = 0.008228 moles O2 per L 

Now apply Avogadro's Number to get the O2 molecules per liter : 
--------------------------------------... 

N sub O2 = ( n sub O2 ) ( N sub AVO ) 

N sub O2 = ( 0.008228 ) ( 6.022 x 10^23 ) 

N sub O2 = 4.96 x 10^21 O2 molecules per liter

It's C.

BlackZzzverrR [31]3 years ago

3 0

C,d,or e you can use the process of elimination to decide...

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A system of ideal gas at 22°C undergoes an ischoric process with an internal energy decrease of 4.30 × 10 3 4.30×103 J to a fina

Komok [63]

Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

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We need to calculate the approximate change in entropy

Using formula of the entropy

$\Delta S=\dfrac{\Delta U}{T}$

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3 years ago

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pav-90 [236]

Answer:

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