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3 years ago

Using a value of Ksp = 1.8 x 10-2 for the reaction PbCl2(s) Pb+2(aq) + 2Cl-(aq).

Chemistry

1 answer:

drek231 [11]3 years ago

7 0

Answer:

Option A and Option B

Explanation:

The concentrations of Pb+2(aq) and Cl-(aq) are expected to decrease (because you have initially very high values of [Pb2+] and [Cl-], they should decrease so that you equilibrium is reestablished, so that Qsp = Ksp).

Solids aren't measured using concentrations but the AMOUNT of PbCl2(s) is expected to increase.

It IS possible to have a system with Ksp > 1.8 x 10^-2 -- if you increase the temperature, then solubility and Ksp are both increased. It is also possible to have a system with Qsp > Ksp (as is true in this case). However, if Qsp > Ksp, the system will shift so that reequilibrium is reestablished. Over time, Qsp will be come equal to Ksp.

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You know that energy cannot be created or destroyed, so what happens to it?

riadik2000 [5.3K]

Mass cannot be created nor destroyed as well.

So, energy just goes into other things.

Example: you are born. You have carbon dioxide in your body, (or star dust). When you die, your body releases that gas.

Make sense?

I hope this helps! (:

6 0

3 years ago

Read 2 more answers

Citrate synthase catalyzes the reaction: ????x????????o????c???????????????????? + ????c????????y???? − ????o???? → c???????????

Nady [450]

The given question is incomplete. The complete question is as follows.

Citrate synthase catalyzes the reaction

Oxaloacetate + acetyl-CoA $\rightarrow$ citrate + HS-CoA

The standard free energy change for the reaction is -31.5 kJ*mol^-1

( a) Calculate the equilibrium constant for this reaction a 37degrees C

Explanation:

(a). It is known that , relation between change in free energy ( $\Delta G$ ) of a reaction and equilibrium constant (K) is as follows.

$\Delta G = -RT \times ln K$

where, T = temperature in Kelvin

The given data is as follows.

T = 310 K, $\Delta G = -31.5 kJ /mol = -31500 J/mol$ (as 1 kJ = 1000 J)

Now, putting the given values into the above formula as follows.

ln K = $\frac{-(\Delta G)}{RT}$

= $\frac{31500}{8.314 \times 310}$

ln K = 12.22

K = antilog (12.22)

= $2.1 \times 10^{5}$

Therefore, we can conclude that value of equilibrium constant for the given reaction is $2.1 \times 10^{5}$ .

6 0

3 years ago

1. When an element has a positive charge :

natita [175]

Answer:

2) AAAAAA

1)BBBBBBBB

3) I think a or c

8 0

3 years ago

Un deportista de 72 kg trepa por una cuerda hasta una altura de 12m. Calcula el incremento de energía potencial gravitatoria que

Oxana [17]

A 72 kg athlete climbs a rope to a height of 12m. Calculate the increase in gravitational potential energy it has experienced.

Answer:

8467.2J

Explanation:

Given parameters:

Mass of the athlete = 72kg

Height of the climb = 12m

Unknown:

Increase in gravitational potential energy it has experienced = ?

Solution:

Gravitational potential energy is the energy due to the position of a body. It is mathematically expressed as;

Gravitational potential energy = m x g x h

m is the mass

g is the acceleration due to gravity = 9.8m/s²

h is the height

Insert the parameters and solve;

Gravitational potential energy = 72 x 9.8 x 12

GPE = 8467.2J

4 0

3 years ago

For a lake in Michigan, researchers have determined that largemouth bass feed on smaller fish, which in turn feed on zooplankton

allsm [11]

Answer:

the expected decline in largemouth bass is 3,000 kg.

Option d) 3,000 kg is the correct answer.

Explanation:

Given the data in the question;

In 2016, there was 600,000 kg of zooplankton in the lake.

In 2017, an accidental runoff of insecticide near the lake caused a 50 percent decline of the zooplankton population in the lake.

Now,

The remaining mass of zooplankton after the 50% decline will be;

⇒ 600,000 kg zooplankton × 50%

⇒ 600,000 × 50/100

⇒ 300,000 kg of zooplankton

Now, with 10 percent trophic efficiency; smaller fish directly feed on zooplankton; decline in smaller fish mass will be;

⇒ 300,000 kg × 10%

⇒ 300,000 × 10/100

⇒ 30000 kg

Finally, with 10 percent trophic efficiency, largemouth bash directly feed on smaller fish, so the expected decline in mass of largemouth bash will be;

⇒ 30000 kg × 10%

⇒ 30000 kg × 10/100

⇒ 3,000 kg

Therefore, the expected decline in largemouth bass is 3,000 kg.

Option d) 3,000 kg is the correct answer.

7 0

3 years ago