Select the appropriate descriptions of the correlation coefficient (r).

The gas co2 is diffusin at steady state through a tube 20 cm long having a diameter of 1 cm and containing n2 at 350 k . the tot

Marrrta [24]

The value is J = 1.71*10^-6 kmol/m^2.s

According to the given conditions we get,

The length the tube l= 0.20m

The diameter of the tube d= 0.01m

The total pressure inside the tube P= 101.32kPa

The partial pressure of CO2 at the first end is

P1= 456mm Hg = 60794.832 Pa

The partial pressure of CO2 at the other end is

P2= 76mm Hg = 10132.472 Pa

The temperature is T = 298 K

The diffusion coefficient D= 1.67* 10^-5 m^2/s

Generally the molar flux of CO2 is mathematically represented as

J= D{p1-p2} / R.T

Here R is the gas constant with value R= 8.314 j/kmol

So we get J= 1.71 * 10^-3 mol/m^2.sec

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4 0

2 years ago

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

So

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

3 years ago

Ayden ran 2 7/8 laps. eduardo ran 1 1/4 laps. how much farther did ayden run than eduardo?

Annette [7]

Answer:

1 5/8

Step-by-step explanation:

2 7/8 - 1 1/4 = 1 5/8.

hope this helps :)

4 0

2 years ago

Read 2 more answers

Sequences.

natta225 [31]

Answer:

192.

Step-by-step explanation:

The formula you can write is y=6x, with x being the sequence number and y being the value of that number. Plugging in 32, the 32nd number would be 192.

8 0

3 years ago

How many pounds of carbon dioxide would you prevent from entering the atmosphere in one year, if you were to replace a 100-watt

VikaD [51]

That's an excellent question but we don't have enough information to answer it. We would need to know how the electrical energy is generated. Nuclear, hydroelectric, coal burning, wind, and natural gas burning all produce different amounts of carbon for the same amount of electrical energy. If the utility that supplies your house uses nuclear or wind, then you can leave both bulbs shining 24/7 and they're not responsible for any carbon.

6 0

3 years ago