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ella [17]
3 years ago
9

All water is located in this layer

Chemistry
1 answer:
11111nata11111 [884]3 years ago
7 0
The troposphere because it is the closest to the earth
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si el calor fuese un fluido ¿en algun momento se agotaria después de que dos objetos se frotaran mutuamente?​
rjkz [21]

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

8 0
2 years ago
3. What are vaccines? Are there different types?​
Inga [223]

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

4 0
3 years ago
Read 2 more answers
Phosphine (PH3) can be prepared by the reaction of calcium phosphide , Ca3P2: based on this equation : Ca3P2 + 6H2O —-&gt; 3 Ca(
MrMuchimi

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>

4 0
2 years ago
What is the role of the greenhouse gases in the atmosphere?
Karolina [17]

Answer:

greenhouse effects are molecules that trap heat on earth atmosphere

7 0
3 years ago
Read 2 more answers
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
nirvana33 [79]

Answer:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]

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:

P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }

With further simplification, we have:

P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]

Then, we have:

P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]

Therefore,

PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]

Using the expansion:

(1-x)^{-1} = 1 + x + x^{2} + ....

Therefore,

PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]

Thus:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]           equation (1)

Using the virial equation of state:

P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]

Thus:

PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]     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[/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

3 0
3 years ago
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