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qwelly [4]
4 years ago
8

A sample of O2 occupies 75 L at 1 atmIf the volume of the sample doubles, what is the new pressure of O 2 ? atm

Chemistry
1 answer:
ipn [44]4 years ago
7 0

Answer:

<em><u>0.5                 if i can have brainliest that would be great</u></em>

Explanation:

P1V1 = P2V2

1 atm x 75L = P2 x 150L

75 atm•L = 150L•P2

Divide both sides by 150L to get P2

75 atm•L / 150L = 150L•P2/150L

0.5 atm = P2

Thus, the new pressure of 02 is 0.5 atmosphere

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Explanation:

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4 years ago
Which orbital do potassium's outermost electrons occupy? 4s 4p 4d 3s
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Potassium outermost electron occupy "4s" orbital
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Read 2 more answers
A very simple assumption for the specific heat of a crystalline solid is that each vibrational mode of the solid acts independen
MatroZZZ [7]

Answer:

4.7 kJ/kmol-K

Explanation:

Using the Debye model the specific heat capacity in kJ/kmol-K

c = 12π⁴Nk(T/θ)³/5

where N = avogadro's number = 6.02 × 10²³ mol⁻¹, k = 1.38 × 10⁻²³ JK⁻¹, T = room temperature = 298 K and θ = Debye temperature = 2219 K  

Substituting these values into c we have

c = 12π⁴Nk(T/θ)³/5  

= 12π⁴(6.02 × 10²³ mol⁻¹)(1.38 × 10⁻²³ JK⁻¹)(298 K/2219 K)³/5

= 9710.83(298 K/2219 K)³/5

= 1942.17(0.1343)³

= 4.704 J/mol-K

= 4.704 × 10⁻³ kJ/10⁻³ kmol-K

= 4.704 kJ/kmol-K

≅ 4.7 kJ/kmol-K

So, the specific heat of diamond in kJ/kmol-K is 4.7 kJ/kmol-K

7 0
3 years ago
A una mezcla de 300g, formada con 60% P/P de Hierro y 40% P/P de Arena, se le adicionan 135g de Cobre y 2,77g de Aluminio. ¿Cuál
Svetllana [295]

Answer:

\%P/P_{hierro}=41.1\%\\\\\%P/P_{arena}=24.4\%\\\\\%P/P_{cobre}=30.8\%\\\\\%P/P_{aluminio}=0.6\%

Explanation:

¡Hola!

En este caso, dado que estamos tratando con problem sobre porcentaje peso/peso de hierro, arena, cobre y aluminio, primero debemos calcular la masa inicial de estos dos primeros en la mezcla original de acuerdo con:

m_{hierro}=300g*0.60=180g\\\\m_{arena}=300*0.40=120g

Ahora si podemos calcular la masa de la mezcla final como la suma de las masas de todos los constituyentes de la mezcla:

m_T=180g+120g+135g+2.77g=437.77g

Finalmente, podemos calcular los porcentajes P/P como se muestra a continuación:

\%P/P_{hierro}=\frac{180g}{437.77g} *100\%=41.1\%\\\\\%P/P_{arena}=\frac{120g}{437.77g} *100\%=24.4\%\\\\\%P/P_{cobre}=\frac{135g}{437.77g} *100\%=30.8\%\\\\\%P/P_{aluminio}=\frac{g}{437.77g} *100\%=0.6\%

¡Saludos!

5 0
3 years ago
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