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Naily [24]
2 years ago
6

7-¿Cuántos moles de CO2 estarán encerrados en un recipiente metálico a una presión de 1.10 Atm. y una temperatura de 31 °C, sabi

endo que el volumen del recipiente es de 15.70 litros?
8- De acuerdo con la ley combinada, dos litros de aire recogidos a 27 °C y 760 mmHg se
llevan a la cima del Pico Duarte donde la temperatura es 12 °C y la presión es 380 mmHg

Cuál es el nuevo volumen de la muestra del aire?​
Chemistry
1 answer:
Aleks [24]2 years ago
4 0

Answer:

7- 0.69 moles.

8- El nuevo volumen es 3.8L.

Explanation:

7- Usando la ley general de los gases:

PV = nRT

PV / RT = n

<em>P es presión = 1.10atm</em>

<em>V es volumen = 15.70L</em>

<em>R es constante de los gases = 0.082atmL/molK</em>

<em>T es temperatura absoluta = 31°C + 273.15K = 304.15K</em>

<em />

Reemplazando:

1.10atm*15.70L / 0.082atmL/molK*304.15K = n

<h3>0.69 moles</h3><h3 />

8- La ley combinada de los gases establece como ecuación:

P1V1 / T1 = P2V2 / T2

<em>Donde P1 es la presión inicial = 760mmHg</em>

<em>V1 volumen inicial = 2L</em>

<em>T1 es temperatura absoluta = 273.15 + 27 = 300.15K</em>

<em>P2 = 380mmHg</em>

<em>T2 = 12°C + 273.15 = 285.15K</em>

<em />

Reemplazando:

760mmHg*2L / 300.15K = 380mmHg*V2 / 285.15K

3.8L = V2

<h3>El nuevo volumen es 3.8L</h3>
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Draw the product formed when cyclohexene is reacted with H2 in the presence of Pt. Note: If adding hydrogen atoms to a carbon at
tigry1 [53]

Answer:

It has been drawn and uploaded as an attachment. Please download it to see the structure.

Explanation:

The product formed as a result of the reaction of cyclohexene with H2​ in presence of Pt (platinum) can be described as catalytic hydrogenation. Catalytic hydrogenation is defined as the process of hydrogen addition in the presence of a catalyst, which in this case is platinum.

Note that Cyclohexene (alkene) is a hydrocarbon molecule represented by the chemical formula, C6​H10​ .

It consists of a double bond. During the hydrogenation reaction, the alkene undergoes an addition reaction to give alkane which is a saturated hydrocarbon as the product.

The first step in order to derive the product is to draw the chemical structure of cyclohexene and identify the double bond present in it.

The final product can be derived by replacing the double bond with the single bond and satisfying all the valences of the carbon atom. The final product structure has been drawn and uploaded as an attachment. Please download it to see the structure.

Ans:

The structure of the cyclohexane thus, formed has been shown as follows with all the hydrogen atoms:

3 0
3 years ago
If the temperature of the water were 0°C and the pressure was above 760 mm Hg, what phase would the water be in?
viva [34]
I would say the answer is gas.
4 0
2 years ago
GIVING BRAINLIEST!!!
vazorg [7]

Answer:

Pecan trees 1 and 4

Explanation:

1: 1 and 4

2a: With good soil and good parent plants

2b: Heredity

6 0
2 years ago
Read 2 more answers
An object has a mass of 6.8 g and volume of 34 mL. What is the density of the object?​
jenyasd209 [6]

Answer:

<h2>Density = 0.2 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3>Density =  \frac{mass}{volume}</h3>

From the question the points are

mass = 6.8 g

volume = 34 mL

Substitute the values into the above formula and solve

That's

<h3>Density =  \frac{6.8}{34}</h3>

We have the final answer as

<h3>Density = 0.2 g/mL</h3>

Hope this helps you

7 0
3 years ago
The rate constant for a certain reaction is measured at two different temperatures:
Talja [164]

Answer: The activation energy Ea for this reaction is 22689.8 J/mol

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

k_1 = rate constant at temperature T_1 = 2.3\times 10^8

k_2 = rate constant at temperature T_2 = 4.8\times 10^8

E_a= activation energy = ?

R= gas constant = 8.314 J/kmol

T_1 = temperature = 280.0^0C=(273+280)=553K

T_2 = temperature = 376.0^0C=(273+376)=649K

Putting in the values ::

ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]

E_a=22689.8J/mol

The activation energy Ea for this reaction is 22689.8 J/mol

3 0
3 years ago
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