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timama [110]
3 years ago
12

4. ¿Cuál es la resistencia de un alambre de

Chemistry
1 answer:
Artemon [7]3 years ago
5 0

Answer:

R=0.0438 Ω

Explicación:

1) Hallar el área o sección del conductor de cobre, usando esta fórmula:

                                      A=π.r² (Pi x radio al cuadrado)

Debido a que conocemos el diámetro (1.5mm) su radio es la mitad de esto es decir 0.75mm, y lo sustituimos en la fórmula:

                                      A=π.(0.75mm)²

                                      A=π(0.5625mm²)

                                      A=1.7671mm²

2) La resistividad del cobre es: rho = 0,0172 y la incluimos en la fórmula siguiente:

                                      R=p  

                                      R=0,0172Ω  x  

Simplificamos:

R=

El resultado es:

R=0.0438 Ω

Explanation:

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3 years ago
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A process stream contains 450 mol/s benzene and 375 mol/s toluene. Calculate the mole fraction of benzene in the process stream.
Tema [17]

Answer:

0.5455

Explanation:

The moles of benzene in the process stream in 1 sec = 450 moles

The moles of toluene in the process stream in 1 sec = 375 moles

So, according to definition of mole fraction:

Mole\ fraction\ of\ benzene=\frac {n_{benzene}}{n_{benzene}+n_{toluene}}

Applying values as:

Mole\ fraction\ of\ benzene=\frac {450}{450+375}

<u>Mole fraction of benzene in the process stream = 0.5455</u>

7 0
4 years ago
Add a temperature of -33.0 Celsius a sample of a confirmed gas eggs or to pressure of 0.5 to 6 ATM if it's volume remains consta
Levart [38]

<u>Answer:</u> The final temperature of the gas is -220.6°C

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=6atm\\T_1=-33^oC=[273-33]K=240K\\P_2=1.31atm\\T_2=?

Putting values in above equation, we get:

\frac{6atm}{240K}=\frac{1.31atm}{T_2}\\\\T_2=\frac{1.31\times 240}{6}=52.4K

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

T(K)=T(^oC)+273

52.4=T(^oC)+273\\\\T(^oC)=-220.6^oC

Hence, the final temperature of the gas is -220.6°C

3 0
4 years ago
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so with every stoichiometry problem with a mass it will make it so you can do the conversion factor with reactants or products.

if you dont understand unit conversions try to study how to set it up. anyways

a.) C12H22O11 has a mass of 342.01 Grams per mole

divide 1.202 G by 342.01 G to get 0.004 miles

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