Answer these two questions, please.

The equilibrium constant Kc for the reaction below is 0.00584 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentra

Vinil7 [7]

Answer:

Explanation:

Given that:

The chemical equation for the reaction is:

Br2(g) ⇌ 2Br(g)

Initially 0.0345M 0.0416M

$Q_C = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416)^2}{(0.0345)}= 0.05016$

$Q_C =0.05016 >>> K_c(0.00584)$

Thus, the given reaction will proceed in the backward direction

The I.C.E table is as follows:

Br2(g) ⇌ 2Br(g)

I 0.0345 0.0416

C +x -2x

E (0.0345+x) (0.0416 -2x)

$K_c = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416-2x)^2}{(0.0345+x)} = 0.00584$

= 0.00173056 - 0.0832x - 0.0832x + 4x² = 0.00584 (0.0345 +x)

= 0.00173056 - 0.166x + 4x² = 2.0148× 10⁻⁴ + 0.00584x

= 0.00173056 - 2.0148× 10⁻⁴ - 0.166x - 0.00584x + 4x²

= 0.00152908 - 0.17184x + 4x²

Solving by using Quadratic formula

x = 0.03038 or 0.0126

For x = 0.03038

At equilibrium

[Br₂] = (0.0345 + 0.03038) = 0.06488 M

[Br] = (0.0416 -2(0.03038)) = - 0.01916 M

Since we have a negative value for [Br], we discard the value for x

For x = 0.0126

At equilibrium

[Br₂] = (0.0345 + 0.0126) = 0.0471 M

[Br] = (0.0416 -2(0.0126)) = 0.0164 M

4 0

3 years ago

Suppose a salt and a glucose solution are separated by a membrane that is permeable to water but not to the solutes. the nacl so

stira [4]

1)

m(NaCl) = 1.95 g
V(H2O) = 250mL
M(NaCl) = 58.5 g/mole

Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:

V(H2O) = 250ml = 250g = 0.25 kg

molality of NaCl:

n(NaCl)=m/M=1.95/58.5= 0.033 mole

molality b(NaCl)=n(NaCl) / V (H2O)= 0.033/0.25 = 0.132 mol/kg

milimolality of NaOH = 0.132/0,001 = 132 mmole/kg

milliosmolality of NaOH = milimolality x N of ions formed in dissociation

Since NaCl dissociates into 2 ions in solution:

milliosmolality of NaOH = 132 x 2 = 264 osmoles/kg

2)

m(gl) = 9 g
V(H2O) = 250mL
M(NaCl) = 180 g/mole

Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:

V(H2O) = 250ml = 250g = 0.25 kg

molality of glucose:

n(gl)=m/M=9/180= 0.05 mole

molality b(gl)=n(gl) / V (H2O)= 0.05/0.25 = 0.2 mol/kg

milimolality of glucose = 0.132/0,001 = 200 mmole/kg

milliosmolality of glucose = milimolality x N of ions formed in dissociation

Since glucose does not dissociate, milimolality and milliosmolality are same:

milliosmolality of glucose = 200 osmoles/kg

3)

The osmosis represents the diffusion of solvent molecules through a semi-permeable membrane that allows passage solvent molecules but does not to the dissolved substance molecule. The osmosis occurs when the concentrations of the solution on both sides of the membrane are different. Since the semi-permeable membrane only permeates the solvent molecules, but not the particles of the dissolved substance, it occurs the solvent diffusion through the membrane, i.e. the solvent molecules pass through the membrane to equalize the concentration on both sides of the membrane. Solvents molecules move from the middle with a lower concentration in the middle with a higher concentration of dissolved substances.

In our case, osmosis will occur because the concentration of NaCl solution and the concentration of glucose solution do not have same values. Osmosis will occur in the direction of glucose solution because it has a lower concentration.

3 0

3 years ago

Which characteristics from the list below cause terrestrial planets to be more likely to have life than Jovian planets?

ddd [48]

Answer:

the Mercury planet is likely to have life but the earth as a lot of oxygen

6 0

3 years ago

A roller coaster starts down a hill at 10 m/s. Three seconds later, its speed is 32 m/s. What is the roller coaster’s accelerati

erica [24]

Final velocity(v) = 32 m/s
Initial velocity(u) = 10 m/s

Using kinematic equation v = u + at,

32 = 10 + a(3)

32-10
--------- = a
3

a = 22
----
3

a = 7.3 m/s^2.

Hence acceleration of the roller coaster is 7.3 m/s^2.

Hope this helps!!

8 0

3 years ago

Calculate the mass of bromine in 50.0 g of potassium bromide

Elodia [21]

Answer:

6 Percent Composition. 1. Molar Mass of KBr K = 1(39.10) = 39.10 Br =1(79.90) =79.90 MM = 119.0 79.90 g 119.0 g = 0.6714 3. 0.6714 x 50.0g = 33.6 g Br 2.

4 0

3 years ago