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OlgaM077 [116]
3 years ago
13

For water at 30C and 1 atm: a= 3.04x10^-4 k^-1 , k =4.52x10^-5 atm ^-1 = 4.46 x10^-10 m^2/N, cpm= 75.3j/9molk), Vm =18.1cm^3/mol

. Find Cv,m of water at 30C and 1 atm
Chemistry
1 answer:
Lapatulllka [165]3 years ago
8 0

Answer:

C_{vm} of water at 30C and 1 atm is 256.834 J/mol·K.

Explanation:

To solve the question, we note the Maxwell relation such as

C_{pm}-C_{vm}=\frac{9T\alpha ^2 V }{K}

Where:

C_{pm} = Specific heat of gas at constant pressure = 75.3 J/mol·K

C_{vm} = Specific heat of gas at constant volume = Required

T = Temperature = 30 °C = 303.15 K

α = Linear expansion coefficient = 3.04 × 10⁻⁴ K⁻¹

K = Volume comprehensibility = 4.52 × 10⁻⁵ atm⁻¹

Therefore,

75.3 - C_v = \frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81  \times 10^{-5}  }{4.52 \times 10^{-5} }

C_{vm}  = \frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81  \times 10^{-5}  }{4.52 \times 10^{-5} } - 75.3 = 256.834 J/mol·K.

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What mass of C6H8O7 should be used every 7.0 X 10^2mg NaHCO3
Savatey [412]

Mass C₆H₈O₇ : 0.531484 g

<h3>Further explanation</h3>

Reaction

3NaHCO₃ (aq) + C₆H₈O₇ (aq) → 3 CO₂ (g) + 3 H₂O (l) + Na₃C₆H₅O₇ (aq)

MW NaHCO₃ : 84 g/mol

mass NaHCO₃ : 7.10² mg=0.7 g

mol NaHCO₃ :

\tt mol=\dfrac{0.7}{84}=0.0083

mol C₆H₈O₇ :

\tt \dfrac{1}{3}\times 0.0083=0.00277

MW C₆H₈O₇ : 192 g/mol

mass C₆H₈O₇ :

\tt mass=0.00277\times 192=0.53184

4 0
2 years ago
What molecules can be made using hydrogen, carbon, and nitrogen?​
Yuki888 [10]

Answer:

Ammonia

Explanation:

Ammonia is the simplest possible molecule made with nitrogen and hydrogen. Methane is the simplest possible molecule made of carbon and hydrogen. Methanol is like methane, but it also has one oxygen atom as well.

8 0
2 years ago
Balance the following chemical equation, then answer the following question.
julsineya [31]
Molar mass:

O2 = 31.99 g/mol
C8H18 = 144.22 g/mol

<span>2 C8H18(g) + 25 O2(g) = 16 CO2(g) + 18 H2O(g)

2 x 144.22 g --------------- 25 x 31.99 g
10.0 g ----------------------?? ( mass of O2)

10.0 x 25 x 31.99 / 2 x 144.22 =

7997.5 / 288.44 => 27.72 g of O2

hope this helps!


</span>

3 0
3 years ago
Calcule la molaridad de una solución acuosa de ácido sulfúrico (H₂SO₄) que se preparó al mezclar, en un recipiente aforado, 4 mo
oee [108]

Considerando la definición de molaridad, la molaridad de una solución acuosa de ácido sulfúrico (H₂SO₄) es 0.5 \frac{moles}{litro}.

La molaridad es una medida de la concentración de un soluto en una disolución que se define como el número de moles de soluto que están disueltos en un determinado volumen.

La molaridad de una solución se calcula dividiendo los moles del soluto por el volumen de la solución:

Molaridad=\frac{numero de moles de soluto}{volumen}

La Molaridad se expresa en las unidades \frac{moles}{litro}.

En este caso, sabes que una solución acuosa se preparó al mezclar 4 moles del ácido con suficiente agua hasta completar 8 litros de solución. Entonces, sabes que:

  • número de moles de soluto= 4 moles
  • volumen= 8 litros

Reemplazando en la definición de molaridad:

Molaridad=\frac{4 moles}{8 litros}

Resolviendo:

Molaridad= 0.5 \frac{moles}{litro}

Finalmente, la molaridad de una solución acuosa de ácido sulfúrico (H₂SO₄) es 0.5 \frac{moles}{litro}.

<em>Aprende más</em>:

  • brainly.com/question/17647411?referrer=searchResults
  • brainly.com/question/21276846?referrer=searchResults

8 0
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