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Ahat [919]
2 years ago
7

4. Juan tiene un cilindro de 40 libras herméticamente cerrado con gas propano y lo conecta mediante una válvula a otro cilindro

de 100 libras completamente vacío. Si Juan abre la válvula del cilindro de 40 dejando pasar el gas al de 100. responda
¿Qué cantidad de gas propano le haría falta al cilindro de 100 y por qué?
Chemistry
1 answer:
ch4aika [34]2 years ago
5 0

Answer:

Ver explicacion

Explanation:

Cuando el cilindro de 40 libras está conectado al cilindro de 100 libras, generamos una presión que continúa hasta que los dos cilindros alcanzan la misma presión de gas en equilibrio. Recuerde que el cilindro de 100 libras estaba inicialmente vacío. Esto significa que su presión inicial es 0. El cilindro de 40 libras ya estaba lleno, por lo que dividimos esta cantidad en dos para tener en cuenta su distribución entre los dos cilindros.

Ahora tenemos 20 libras de gas propano presentes en cada cilindro. La implicación de esto es que, en el cilindro de 100 libras, necesitamos 80 libras adicionales para completar las 100 libras.

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The element that is a nonmetal, a gas, and has an element symbol that starts with the letter a is Ar (Argon).
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Uranium-235 decays to form lead-207. The half-life of uranium-235 is 704,000,000 years. What is the approximate age of an igneou
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Answer:

The approximate age of the igneous rock is 7.042\times 10^{6} years.

Explanation:

Chemically speaking, the Uranium-235 decays to form Lead-207 throughout time. All isotopes decay exponentially by means of the following model:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} } (1)

In addition, we can determine the time constant of the former isotope in terms of its half life, that is:

\tau = \frac{t_{1/2}}{\ln 2} (2)

Where:

m_{o} - Initial mass of Uranium-235, in atoms.

m(t) - Current mass of Uranium-235, in atoms.

\tau - TIme constant, in years.

t_{1/2} - Half-life of Uranium-235, in years.

Please remind that a 1 : 1 ratio of Uranium-235 to Lead-207 atoms means that current mass of Uranium-235 is a half of its initial mass. If we know that m(t) = 0.5\cdot m_{o} and t_{1/2} = 7.04\times 10^{8}\,yr, then the approximate age of the igneous rock is:

\tau = \frac{7.04\times 10^{8}\,yr}{\ln 2}

\tau \approx 1.016\times 10^{9}\,yr

t = -\tau \cdot  \ln \left(\frac{m(t)}{m_{o}} \right)

t = -(1.016\times 10^{9}\,yr)\cdot \ln \frac{1}{2}

t \approx 7.042\times 10^{6}\,yr

The approximate age of the igneous rock is 7.042\times 10^{6} years.

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Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. Classify each of the following me
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Answer:

a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

Explanation:

Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. The reactivity series of metals arranges metals based on decreasing order of reactivity. The more reactive metals are found higher up in the series while the least reactive metals are found at the lower ends of the series. Thus, metals above iron in the reactivity series can serve as sacrificial anodes by protecting against corrosion, while those lower than iron cannot.

Based on the reactivity series, the following metals can be classified as either a sacrificial anode for iron or not:

a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

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Force = Mass * Acceleration = 59 kg * 9.75 m/s^2 = 575.25 N
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