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baherus [9]
3 years ago
13

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

Physics
2 answers:
hram777 [196]3 years ago
6 0
N sub air = ( P ) ( V ) / ( R ) ( T ) 

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

<span>n sub air = 0.03928 mol per L </span>

<span>R = universal gas constant = 0.08205 atm - L per mol - deg K </span>

<span>Find moles O2 per liter : </span>
<span>--------------------------------------... </span>

<span>n sub O2 = ( y sub O2 ) ( n sub air ) </span>

<span>n sub O2 = ( 0.20948 ) ( 0.03928 ) </span>

<span>n sub O2 = 0.008228 moles O2 per L </span>

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

<span>N sub O2 = ( n sub O2 ) ( N sub AVO ) </span>

<span>N sub O2 = ( 0.008228 ) ( 6.022 x 10^23 ) </span>

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

It's C.

</span>
BlackZzzverrR [31]3 years ago
3 0
C,d,or e you can use the process of elimination to decide...
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Which part(s) of the electromagnetic spectrum are visible to humans?
jarptica [38.1K]
Color aka the visible light spectrum
4 0
3 years ago
calculate your power in watts if you do 500 N*m of work picking up a dumbbell,and takes 5 s to perform that task?​
Slav-nsk [51]

Answer:

100watts

Explanation:

Given parameters:

Workdone  = 500Nm

Time taken = 5s

Unknown:

Power in watts = ?

Solution:

Power is the rate at which work is done;

   Power  =\frac{workdone }{time}

Input the parameters and solve;

   Power = \frac{500}{5}  = 100watts

6 0
3 years ago
Sobre un gas contenido en un cilindro provisto de un pistón se realiza un trabajo de 7000 Joules, mediante un proceso isotérmico
natita [175]

Answer:

En un proceso isotérmico, es decir, la temperatura no cambia, el trabajo puede escribirse como:

W = n*R*T*Ln(P1/P2)

Donde P1 es la presión inicial y P2 la presión final.

Donde las cantidades:

n =  número de moles

R = constante de los gases ideales

T = temperatura no cambian.

Y sabemos que la ecuación de la energía interna es:

U = C*n*R*T

Donde C es otra constante que depende del gas.

De aca, podemos concluir que ninguna de estas variables cambia en nuestro proceso, por lo que la variación de la energía interna es cero.

U2 - U1 = 0

b) Para el calor cedido o absorbido, la formula básica es:

ΔQ = C*(T2 - T1)

Donde ΔQ es el calor absorbido o cedido por el gas, C es una constante que depende del gas, T2 es la temperatura final del gas y T1 es la temperatura inicial del gas.

Como la temperatura no cambia en el proceso, entonces:

T2 = T1

ΔQ = C*(T2 - T1) = C*0 = 0

No hay calor absorbido ni cedido.

c) Podemos concluir que en un proceso isotérmico la energía interna no cambia, y no hay un intercambio de calor.

8 0
3 years ago
The tides are...
Nadusha1986 [10]

well i think that

a constant energy source

5 0
3 years ago
Read 2 more answers
Sorry<br>Please ignore this idk how to delete it
MissTica

Answer:

lol hahahaha okie dokie

3 0
3 years ago
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