Jupiter revolves around the Sun in a predictable pattern because

Consider the following system at equilibrium: 2A(aq)+2B(aq)⇌5C(aq) Classify each of the following actions by whether it causes a

Alexandra [31]

Answer:

Rightwardshift: (a), (b), (f) and (h)

Leftwardshift: (c), (d), and (e)

No shift: (g)

Explanation:

1. Balanced chemical equation (given):

$2A(aq)+2B(aq)\rightleftharpoons 5C(aq)$

2. Equilibrium constant

The equilibrium constant is the ratio of the product of the concentrations of the products, at equilibrium, each raised to its stoichiometric coefficient, to the product of the concentrations of the reactants, at the equilibrium, each raised to its stoichiometric coefficient.

$K_{c}=\frac{[C(aq)]^5}{[A(aq)]^2\cdot [B(aq)]^2}$

a. Increase [B]

Rightward shift

Since, by assumption, the temperature of the reaction is the same, the equilibrium constant $K_{c}$ is the same, meaning that an increase in the concentration of the species B must cause a rightward shift to increase the concentration of the species C, such that the ratio expressed by the equilibrium constant remains unchanged.

b. Increase [A]

Rightward shift.

This is exactly the same case for the increase of [B], since it is in the same side of the equilibrium chemical equation.

c. Increase [C]

Leftward shift.

C is on the right side of the equilibrium equation, thus, following Le Chatelier's principle, an increase of its concentration must shift the reaction to the left to restore the equilibrium. Of course, same conclusion is drawn by analyzing the expression for $K_{c}$ : by increasing the denominator the numerator has to increase to keep the same value of $K_{c}$

d. Decrease [A]

Leftward shift.

This is the opposite change to the case {b), thus it will cause the opposite effect.

e. Decrease[B]

Leftward shift.

This is the opposite to case (a), thus it will cause the opposite change.

f. Decrease [C]

Rightward shift.

This is the opposite to case (c), thus it will cause the opposite change.

g. Double [A] and reduce [B] to one half

No shift

You need to perform some calculations and determine the reaction coefficient, $Q_c$ to compare with the equilibrium constant $K_{c}$ .

The expression for $Q_c$ has the same form of the equation for $K_{c}$ . but the it uses the inital concentrations instead of the equilibrium concentrations.

$Q{c}=\frac{[C(aq)]^5}{[A(aq)]^2\cdot [B(aq)]^2}$

Doubling [A] and reducing [B] to one half would leave the product of [A]² by [B]² unchanged, thus $Q_c$ will be equal to $K_{c}$ .

When $Q_c$ = $K_{c}$ the reaction is at equilibrium, so no shift will occur.

h. Double both [B] and [C]

Rightward shift.

Again, using the expression for $Q_c$ , you will realize that the [C] is raised to the fifth power (5) while [B] is squared (power 2). That means that $Q_c$ will be greater than $K_{c}$ ..

When $Q_c$ > $K_{c}$ the equilibrium must be displaced to the left some of the reactants will need to become products, causing the reaction to shift to the right.

Summarizing:

Rightwardshift: (a), (b), (f) and (h)

Leftwardshift: (c), (d), and (e)

No shift: (g)

4 0

4 years ago

The chemical equation below shows the reaction between tin (Sn) and hydrogen fluoride (HF). Sn + 2HF mc017-1.jpg SnF2 + H2 The m

Sergio [31]

Tin and Hydrofluoric Acid reacts as shown,

 Sn + 2 HF → SnF₂ + H₂

According to Equation,

40.02 g (2 Moles) HF Required = 1 Mole of Sn for complete Reaction

So,
40 g of HF will require = X Moles of Sn

Solving for X,
X = (40 g × 1 Mole) ÷ 40.02 g

X = 0.999 Moles ≈ 1 Mole
Result:
40 g of HF requires 1 Mole of Tin (Sn) for complete Reaction to produce SnF₂ and H₂.

7 0

3 years ago

Read 2 more answers

Suppose you have designed a new thermometer called the x thermometer. On the x scale the boiling point of water is 129 ?x and th

djyliett [7]

(x1,y1) (x2,y2)

(32,15) (212,135)

y = mx + b

m=(y2-y1)/(x2-x1)

m=.66667

y = .66667*(x) + b

Plug in #s to test for b value

15 = .66667*(32) + b

b = -6.334

y = mx + b

y = .6667*(x) + (-6.334)

F = .6667*(F) + (-6.334)

F = (.6667*(F) = -6.334

(F - .6667F) = -6.334

F(1 - .6667) = -6.334

F = -19

ThermX = F @ -19 degrees

Brainliest please!

3 0

3 years ago

what is the final molarity of a dilute solution of NaOH that is prepared by pipetting 5.00 mL of 7.00 M NaOH into a 100mL volume

Angelina_Jolie [31]

Answer:

$M_{diluted}=0.35M$

Explanation:

Hello,

In this case, since in a dilution process the moles of the solute must remain unchanged, we use the volumes and molarities as shown below:

$M_{diluted}V_{diluted}=M_{concentrated}V_{concentrated}$

Clearly, the concentrated solution is 7.00M and the diluted solution is unknown, thus, the concentration of the diluted solution after the dilution to 100 mL is:

$M_{diluted}=\frac{M_{concentrated}V_{concentrated}}{V_{diluted}} =\frac{7.00M*5.00mL}{100mL}\\ \\M_{diluted}=0.35M$

Best regards.

4 0

4 years ago

A 0.227 mol chunk of dry ice (solid CO2) changes to gas. What is the volume of that gas measured at 27 °C and 740 mmHg?

jeka57 [31]

Answer:

Explanation:3.2 ft 2 fti2 ft 4 ft ft2

4 0

3 years ago