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2 years ago

14

When does the net flux of dissolved molecules stop

1 answer:

Mnenie [13.5K]2 years ago

5 0

The net fluid flux usually stops when the amount or concentration of osmotic active molecules is equal (the same) in the two areas (compartment)

For better understanding let's explain what statement above means

Net fluid flux simply ends when the rate(concentration) of the substance is at the same proportion in both areas. It does not implies that movement is not taken place by molecules of that substance but that there is no final movement in one direction simply pit it that the molecules of substance are equally going in both directions (area).

From the above we can therefore say that the answer The net fluid flux usually stops when the amount or concentration of osmotic active molecules is equal (the same) in the two areas (compartments) correct

Learn more about osmotic pressure from:

brainly.com/question/15180636

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A 2.0 kg tumor is being irradiated by a radioactive source.The

navik [9.2K]

Answer:

426136363636.36365 Bq

Explanation:

Grays = 12

Mass = 2 kg

Time = 880

Energy = 0.4 MeV

Activity is given by

$Activity=\dfrac{Grays\times Mass}{Time\times 0.4}\\\Rightarrow Activity=\dfrac{12\times 2}{880\times 0.4\times 1.6\times 10^{-13}}\\\Rightarrow Activity=426136363636.36365\ Bq$

The activity N//t of the radioactive source is 426136363636.36365 Bq

6 0

4 years ago

What can be proved from the given figure?

Neko [114]

It prove that liquid pressure increases with depth

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3 years ago

A 7.0 mm -diameter copper ball is charged to 40 nC. What fraction of its electrons have been removed? The density of copper is 8

mylen [45]

Answer:

$f = 2.6 \times 10^{-13}$

Explanation:

Let the mass of copper ball is "m" gram

now the total number of copper atom present in the ball is given as

$N = \frac{m}{29} \times 6.02 \times 10^{23}$

now the total number of electrons in one copper atom is 29

so total number of electrons in given sample of copper ball is

$N_e = m(6.02 \times 10^{29})$

now diameter of the ball is 7.0 mm

density of the ball = $8900 kg/m^3$

now we have

$m = (\frac{4}{3}\pi r^3)(8900)$

$m = (\frac{4}{3}\pi(\frac{0.007}{2})^3)(8900)$

$m = 1.6 gram$

now we have

$N_e = 9.63 \times 10^{23}$

now the charge on the copper ball is 40 nC

so the number of electrons removed

$Q = ne$

$40 \times 10^{-9} = n(1.6 \times 10^{-19}$

$n = 2.5 \times 10^{11}$

so the fraction of number of electrons removed is given as

$f = \frac{n}{N_e}$

$f = \frac{2.5 \times 10^{11}}{9.63 \times 10^{23}}$

$f = 2.6 \times 10^{-13}$

7 0

4 years ago

A particle is confined to a one-dimensional box that is 50 pm long. What is the smallest possible uncertainty in momentum for th

nevsk [136]

Answer:

The smallest possibility is 0.01E-22kgm/s

Explanation:

Using

Momentum= h/4πx

= 6.6x 10^-34Js/ 4(3.142* 50*10-12m)

= 0.01*10^-22kgm/s

6 0

3 years ago

An electric device delivers a current of 5.0 A to a circuit. How many electrons flow through this circuit in 5 s?

IgorC [24]

Answer:

1.6×10²⁰

Explanation:

An ampere is a Coulomb per second.

1 A = 1 C / s

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5.0 A × 5 s = 25 C

The number of electrons is:

25 C × (1 electron / 1.6×10⁻¹⁹ C) = 1.6×10²⁰ electrons

7 0

4 years ago