50 points plz help

Compare and contrast diffusion and osmosis. Include at least one similarity and one difference between the two processes.

Softa [21]

Osmosis is the diffusion of water across a semipermeable membrane (usually cell membrane) from a region of low solute concentration to a more concentrated solution so it can reach equilibrium (balance).

Diffusion is a spontaneous movement of particles from an area of high concentration to an area of low concentration.

Both results in particles moving and help balance out the concentrations.
Also, in osmosis, the water molecules are moving. In diffusion, it is the solutes moving.

I hope this helps and explains well.

4 0

3 years ago

Dinitrogen trisulfide symbol

notka56 [123]

The symbol for dinitrogen trisulfide is N2S3

4 0

3 years ago

Read 2 more answers

Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

Answer: The equilibrium concentration of water is 0.597 M

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

3 years ago

How many potassium atoms are in 250.0 grams of potassium?

loris [4]

Hi!

The correct option would be 3.85x10^(24)

To find the number of atoms in 250g of potassium, we need to first calculate the number of atoms in

1 mole of Potassium = 39g which contains 6.022x10^(23) atoms of K

(Avogadro's constant value for the amount of molecules/atoms in one mole of any substance)

Solution

So as 39g of Potassium contains 6.022x10^(23) K atoms

1g of Potassium would contain 6.022x10^(23) / 39 = 1.544 x10^(22) atoms

So 250g of Potassium would contain 1.544x10^(22) x 250 = 3.86x10^(24) atoms

3 0

3 years ago

If the initial temperature of a movable cylinder was 50 degrees Celsius

slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

5 0

3 years ago