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3 years ago

How many dots would go around xenon in electron dot diagram

Chemistry

1 answer:

ratelena [41]3 years ago

4 0

Answer:

Explanation:

Since Xe (xenon) is a noble gas, it has 8 valence electrons. It's outer shell is full and it is a stable element and doesn't like to react. Therefore, since the electron dot diagram is about valence electrons, 8 would be your answer.

You might be interested in

Where would you find water on the ph scale?

vladimir2022 [97]

Answer: between 6.5 and 8.5

Explanation:

As shown in my science book it appears in the ph scale

5 0

2 years ago

A 10.0 mL sample of 0.75 M CH3CH2COOH(aq) is titrated with 0.30 M NaOH(aq) (adding NaOH to CH3CH2COOH). Determine which region o

Llana [10]

Answer:

5.75

Explanation:

Weak acid and strong base,

Make it a two part problem

The first will be Partial neutralization of the weak acid, while the second is the equilibrium and final pH

1.Parameters

HC2H3O2 =10mL x 0.75M = 7.5 mmol

NaOH =22mL x 0.30M = 6.6 mmol

R HC2H3O2 + OH- --> C2H3O2- + H2O

I 7.5 mmol 6.6 mmol 0 mmol ignore

C -6.6 mmol -6.6 mmol +6.6 mmol

E 0.9 mmol 0.0 mmol 6.6 mmol

2.) HC2H3O2 --> H+ + C2H3O2-

Initial concentrations

[HC2H3O2]: 0.9 mmol / 32mL = 0.0281M

[H+]:

[C2H3O2-]: 6.6 mmol x 32mL = 0.206M

Concentrations at equilibrium

[HC2H3O2]: 0.0281M - x

[H+]: [C2H3O2-]: 0.206M + x

If x is small and can be ignored except [H+]

Substitute and solve

1.3 x 10-5 = (x)(0.206)

(0.0281)

X= 1.77 x 10-6M => pH = 5.75

3 0

3 years ago

Determine the volume of carbon dioxide gas produced at STP when 2.5 g of nitroglycerin decomposes by this balanced equation. 4C3

PolarNik [594]

Answer:

0.67 L

Explanation:

Given data:

Volume of carbon dioxide = ?

Mass of nitroglycerine = 2.5 g

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Solution:

Chemical equation:

4C₃H₅N₃O₉ → 12CO₂ + 10 H₂O + 6N₂ + O₂

Number of moles of nitroglycerine:

Number of moles = mass/ molar mass

Number of moles = 2.4 g/ 227.1 g/mol

Number of moles = 0.01 mol

Now we will compare the moles of nitroglycerine and carbon dioxide from balance chemical equation.

C₃H₅N₃O₉ : CO₂

4 : 12

0.01 : 12/4×0.01 = 0.03 mol

Volume of CO₂:

PV = nRT

V = nRT/P

V = 0.03 mol× 0.0821 atm.L/mol.K× 273.15 K / 1 atm

V = 0.67 L

4 0

3 years ago

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

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide:You add 40.7 g of M

Alex787 [66]

: 4 HCl + MnO2 → MnCl2 + 2 H2O + Cl2

(33.7 g MnO2) / (86.93691 g MnO2/mol) = 0.38764 mol MnO2
(45.3 g HCl) / (36.4611 g HCl/mol) = 1.2424 mol HCl

(a)
1.2424 moles of HCl would react completely with 1.2424 x (1/4) = 0.3106 mole of MnO2, but there is more MnO2 present than that, so MnO2 is in excess and HCl is the limiting reactant.

(b)
(1.2424 mol HCl) x (1 mol Cl2 / 4 mol HCl) x (70.9064 g Cl2/mol) = 22.0 g Cl2

(c)
(77.7% of 22.0 g Cl2) = 17.1 g Cl2

5 0

3 years ago