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Lubov Fominskaja [6]
3 years ago
12

Si en un planeta la presión atmosférica es mayor que 5,1 atm terrestres, los habitantes podrían nadar en ríos de dióxido de carb

ono líquido?
Chemistry
1 answer:
Sphinxa [80]3 years ago
5 0

Answer:

Los habitantes del planeta con una atmósfera superior a 5,1 atm de la Tierra, no estarían nadando en ríos de dióxido de carbono líquido

Explanation:

De las tablas de datos termodinámicos, la presión a la que el vapor de dióxido de carbono está en equilibrio con su estado líquido a una temperatura ambiente de 25 ° C es 6,401 kPa, lo que equivale a 63,17296 atm.

Por lo tanto, a una presión de 5.1 de la atmósfera terrestre, el dióxido de carbono es completamente gaseoso y los habitantes del planeta con una presión atmosférica de 5.1 atm de la Tierra todavía observarían solo hidrógeno gaseoso y no estarían nadando en ríos de dióxido de carbono líquido.

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can someone help me more clearly understand the difference between heterogeneous and homogeneous? Give an example of each plz.
const2013 [10]
Heterogenous mixtures are unevenly mixed. Like oil and vinegar in vinaigrette if it is not emulsified well enough and they separate. Any case where two things are not evenly distributed within each other.

Homogenous mixtures are evenly mixed throughout. Like salt water or kool-aid (when it's mixed).

Hope this helps!
3 0
3 years ago
The decomposition of ammonia is: 2 NH3(g) = N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammoni
hjlf

Answer:

A) = 4.7 × 10⁻⁴atm

Explanation:

Given that,

Kp = 1.5*10³ at 400°C

partial pressure pN2 = 0.10 atm

partial pressure pH2 = 0.15 atm

To determine:

Partial pressure pNH3 at equilibrium

The decomposition reaction is:-

2NH3(g) ↔N2(g) + 3H2(g)

Kp = [pH2]³[pN2]/[pNH3]²

pNH3 =√ [(pH2)³(pN2)/Kp]

pNH3 = √(0.15)³(0.10)/1.5*10³ = 4.74*10⁻⁴ atm

K_p = \frac{[pH_2] ^3[pN_2]}{[pNH_3]^2} \\pNH_3 = \sqrt{\frac{(pH_2)^3(pN_2)}{pNH_3} } \\pNH_3 = \sqrt{\frac{(0.15)^3(0.10)}{1.5 \times 10^3} } \\=4.74 \times 10^-^4atm

= 4.7 × 10⁻⁴atm

4 0
3 years ago
Read 2 more answers
There are 31.1 grams (g) in 1 troy ounce (ozt). How many ozt of gold are extracted from 1 short ton of average Nevada ore
devlian [24]

This problem is providing us with the mass equivalent to one troy ounce. Thus, the troy ounces of gold in one short ton of average Nevada ore is required and found to be the 0.103 otz according to the following dimensional analysis.

<h3>Dimensional analysis</h3>

In chemistry, a raft of problems do not always provide an equation in order to be solved yet dimensional analysis can be applied, so as to obtain the desired amount in the required units.

Thus, since this problem asks for try ounces in an average Nevada ore,  which has 3.2 grams of gold per short ton of ore, one can solve the following setup in order to obtain the required answer in otz:

\frac{3.2gAu}{1short-ton}*1short-ton*\frac{1otz}{31.1g} \\

Where the short tons are cancelled out as well as the grams, in order to obtain:

0.103 otz

Learn more about dimensional analysis: brainly.com/question/10874167

4 0
2 years ago
Solve the question that follows using the equation for the conversion of Celsius to Fahrenheit. F=95(C)+32 On February 9, 1934,
ahrayia [7]

Answer:

Buffalo, NY was colder on February 9, 1934

Explanation:

Temperature is the measure of thermal energy present in a given substance. It is a physical property of a matter which can be measured on the Celsius scale (°C), Fahrenheit scale (°F), and Kelvin scale (K).

<u>To convert Celsius (°C) to Fahrenheit (°F), we use the equation:</u>

°F = 9/5 (°C) + 32

Given: On February 9, 1934,

Temperature in Buffalo, NY: T₁ = (- 29) °C

Temperature in Anchorage, AL: T₂ = 19 °F

<u>To convert T₁ from °C to °F:</u>

9/5 (- 29 °C) + 32 = (- 52.2) + 32 = (- 20.2) °F

So,

Temperature in Buffalo, NY: T₁ = (- 20.2) °F

Temperature in Anchorage, AL: T₂ = 19 °F

⇒ T₁ < T₂

<u>Therefore, Buffalo, NY was colder on February 9, 1934.</u>

4 0
4 years ago
Micronutrients can be classified based on the presence of ______ in their chemical structures. According to this classification
natali 33 [55]

Answer:

Carbon

Explanation:

Micronutrients include elements or groups needed by the body in order to survive. These elements or groups include vitamins and minerals.  Micronutrients can be divided into four categories:

1. water-soluble vitamins

2. Fat-soluble vitamins

3. Macrominerals

4. Trace minerals

8 0
3 years ago
Read 2 more answers
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