Si el calor fuese un fluido no se agotaría, solamente cambiaría su forma o localización.
<h3>¿Qué ocurre cuando dos objetos con diferente temperatura están en contacto?</h3>
Cuando dos objetos de diferente temperatura están en contacto el calor fluye del objeto con mayor temperatura o más caliente, al objeto con menor temperatura o más frío. Esto ocurre hasta que ambos objetos alcanzan la misma temperatura.
<h3>¿Qué sucede con el calor?</h3>
El calor o energía termina se disipa hacia el entorno o la atmosfera cuando un cuerpo se enfría o hacia otro cuerpo si hay transferencia de calor, sin embargo, el calor no desaparece debido a la ley de la conservación de la energía y materia.
<h3>¿Qué sucedería si el calor fuera un fluido?</h3>
Si el calor fuera un fluido este fluido no desaparecería ni se agotaría, solamente se podría disipar a la atmósfera o ambiente.
Aprenda más sobre calor en: brainly.com/question/15890992
Answer:Four types of vaccines are currently available: Live virus vaccines use the weakened (attenuated) form of the virus. The measles, mumps, and rubella (MMR) vaccine and the varicella (chickenpox) vaccine are examples
Taking into account the reaction stoichiometry, you can observe that:
- one mole of Ca₃P₂ produces 2 mol of PH₃.
- the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
Ca₃P₂ + 6 H₂O → 3 Ca(OH)₂ + 2 PH₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Ca₃P₂:1 mole
- H₂O: 6 moles
- Ca(OH)₂: 3 moles
- PH₃: 2 moles
The molar mass of the compounds is:
- Ca₃P₂: 182 g/mole
- H₂O: 18 g/mole
- Ca(OH)₂: 74 g/mole
- PH₃: 34 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Ca₃P₂: 1 mole ×182 g/mole= 182 grams
- H₂O: 6 moles× 18 g/mole= 108 grams
- Ca(OH)₂: 3 moles ×74 g/mole= 222 grams
- PH₃: 2 moles ×34 g/mole= 68 grams
<h3>Correct statements</h3>
Then, by reaction stoichiometry, you can observe that:
- one mole of Ca₃P₂ produces 2 mol of PH₃.
- the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.
Learn more about the reaction stoichiometry:
<u>brainly.com/question/24741074</u>
<u>brainly.com/question/24653699</u>
Answer:
greenhouse effects are molecules that trap heat on earth atmosphere
Answer:
![PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B1%20%2B%20%28b-%5Cfrac%7Ba%7D%7BRT%7D%29%5Cfrac%7B1%7D%7BV_%7Bm%7D%20%7D%20%2B%20%5Cfrac%7Bb%5E%7B2%7D%20%7D%7BV%5E%7B2%7D%20_%7Bm%7D%20%7D%20%2B%20...%5D)
B = b -a/RT
C = b^2
a = 1.263 atm*L^2/mol^2
b = 0.03464 L/mol
Explanation:
In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:
Using the van deer Waals equation of state:
With further simplification, we have:
![P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]](https://tex.z-dn.net/?f=P%20%3D%20RT%5B%5Cfrac%7B1%7D%7BV_%7Bm%7D-b%20%7D%20-%20%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%5E%7B2%7D%20%7D%5D)
Then, we have:
![P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BRT%7D%7BV_%7Bm%7D%20%7D%20%5B%5Cfrac%7B1%7D%7B1-%5Cfrac%7Bb%7D%7BV_%7Bm%7D%20%7D%20%7D%20-%20%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%7D%5D)
Therefore,
![PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B%281-%5Cfrac%7Bb%7D%7BV_%7Bm%7D%20%7D%29%20%5E%7B-1%7D%20-%20%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%7D%5D)
Using the expansion:
Therefore,
![PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B1%2B%5Cfrac%7Bb%7D%7BV_%7Bm%7D%20%7D%2B%5Cfrac%7Bb%5E%7B2%7D%20%7D%7BV_%7Bm%7D%20%5E%7B2%7D%20%7D%20%2B%20...%20-%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%7D%5D)
Thus:
equation (1)
Using the virial equation of state:
![P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]](https://tex.z-dn.net/?f=P%20%3D%20RT%5B%5Cfrac%7B1%7D%7BV_%7Bm%7D%20%7D%2B%20%5Cfrac%7BB%7D%7BV_%7Bm%7D%20%5E%7B2%7D%7D%2B%5Cfrac%7BC%7D%7BV_%7Bm%7D%20%5E%7B3%7D%20%7D%2B%20...%5D)
Thus:
![PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B1%2B%20%5Cfrac%7BB%7D%7BV_%7Bm%7D%20%7D%2B%20%5Cfrac%7BC%7D%7BV_%7Bm%7D%20%5E%7B2%7D%20%7D%20%2B%20...%5D)
equation (2)
Comparing equations (1) and (2), we have:
B = b -a/RT
C = b^2
Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.
![b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol](https://tex.z-dn.net/?f=b%20%3D%20%5Csqrt%7BC%7D%20%3D%20%5Csqrt%7B1200%7D%20%3D%2034.64%5Btex%5Dcm%5E%7B3%7D%2Fmol)
[/tex] = 0.03464 L/mol
a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2
