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2 years ago

14

Be sure to answer all parts. Write a balanced equation and Kb expression for the following Brønsted-Lowry base in water: benzoat

e ion, C6H5COO−. Include the states of all reactants and products in your equation. You do not need to include states in the equilibrium expression. Balanced equation: ⇌ Kb expression:

1 answer:

Cerrena [4.2K]2 years ago

3 0

Answer:

The balanced reaction is:-

$C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}$

$K_b$ expression is:-

$K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}$

Explanation:

Benzoate ion is the conjugate base of the benzoic acid. It is a Bronsted-Lowry base and the dissociation of benzoate ion can be shown as:-

$C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}$

The expression for dissociation constant of benzoate ion is:

$K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}$

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Pachacha [2.7K]

Problem One (left)

This is just a straight mc deltaT question

<em><u>Givens</u></em>

m = 535 grams

c = 0.486 J/gm

tf = 50

ti = 1230

Formula

E = m * c * (ti - tf)

Solution

E = 535 * 0.486 * ( 1230 - 50)

E = 535 * 0.486 * (1180)

E = 301077

Answer: A

Problem Two

This one just requires that you multiply the two numbers together and cut it down to 3 sig digits.

E = H m

H = 2257 J/gram

m = 11.2 grams

E = 2257 * 11.2

E = 25278 to three digits is 25300 Joules. Anyway it is the last one.

Three

D and E are both incorrect for the same reason. The sun and stars don't contain an awful lot of Uranium (1 part of a trillion hydrogen atoms). It's too rare. The other answers can all be eliminated because U 235 is pretty stable in its natural state. It has a high activation complex.

Your best chance would be enriched Uranium (which is another way of saying refined uranium). That would be the right environment. Atomic weapons and nuclear power plants (most) used enriched Uranium. You can google "Little Boy" if you want to know more.

Answer: B

Four

The best way to think about this question is just to get the answer. Answer C.

A: incorrect. Anything sticking together implies a larger and larger result. Gases don't work that way. They move about randomly.

B: Wrong. Heat and Temperature especially depend on movement. Stopping is not permitted. If a substance's molecules stopped, the substance would experience an extremely uncomfortable temperature drop.

C: is correct because the molecules neither stop nor do they stick. The hit and move on.

D: Wrong. An ax splitting something? That is not what happens normally and not with ordinary gases. It takes more energy that mere collisions or normal temperatures would provide to get a gas to split apart.

E: Wrong. Same sort of comment as D. Splitting is not the way these things work. They bounce away as in C.

Five

Half life number 1 would leave 0.5 grams behind.

Half life number 2 would leave 1/2 of 1/2 or 1/4 of the number of grams left.

Answer: 0.25

Answer C

6 0

2 years ago

6. Under standard-state conditions, what spontaneous reaction will occur in aqueous solution among the ions Ce4+, Ce3+, Fe3+, an

xz_007 [3.2K]

Answer:

ΔG° = -80.9 KJ

Assuming this reaction takes place at room temperature (25 °C):

K= $1.53x10^{14}$

Explanation:

1) Reduction potentials

First of all one should look up the reduction potentials for the species envolved:

$Ce^{4+} + e$ → $Ce^{3+}$ E°red=1.61V

$Fe^{3+} + e$ → $Fe^{2+}$ E°red=0.771V

2) Redox pair

Knowing their reduction pontentials one can determine a redox pair: one species must oxidate while the other is reducing. <u>Remember: the table gives us the reduction potential, so if we want to know the oxidation potential all that has to be done is reverce the equation and change the potencial signal (multiply to -1).</u>

1) $Ce^{4+}$ reduces while $Fe^{2+}$ oxidates

(oxidation) $Fe^{2+}$ → $Fe^{3+} + e$ E°oxi=-0.771V

(reduction) $Ce^{4+} + e$ → $Ce^{3+}$ E°red=1.61V

(overall equation) $Fe^{2+}+Ce^{4+}$ → $Ce^{3+}+Fe^{3+}$ E°=Ereduction + Eoxidation= 1.61 v+(-0.771 v) = 0.839v

The cell potential can also be calculated as the cathode potencial minus the anode potential:

E° = E cathode - E anode =1.61 v - 0.771 v=0.839 v

3) Gibbs free energy and Equilibrium constant

ΔG°=-nFE°, where 'n' is the number of electrons involved in the redox equation, in this case n is 1. 'F' is the Faraday constant, whtch is 96500 C. E° is the standard cell potencial.

ΔG°=-nFE°=-1*96500*0.839

ΔG° = - 80963 J = -80.9 KJ

The Nerst equation gives us the relation of chemical equilibrium and Electric potential.

E=E°- $\frac{RT}{nF} Ln Q$

Where 'R' is the molar gas constant (8.314 J/mol)

It's known that in the equilibrium E=0, so the Nerst equation, at equilibrium, becomes:

E°= $\frac{RT}{nF} Ln K$

Isolating for 'K' gives:

$K=e^{\frac{nFE^{o} }{RT} }$

This shows that 'K' is a fuction of temperature. Assuming this reaction takes place at room temperature (25 °C):

K= $1.53x10^{14}$

6 0

2 years ago

Calculate the Lattice Energy of KCl(s) given the following data using the Born-Haber cycle: ΔHsublimation K = 79.2 kJ/mol IE1 K

DanielleElmas [232]

Answer:

Explanation:

The following equation relates to Born-Haber cycle

$\triangle H_f = S + \frac{1}{2} B + IE_M - EA_X+ U_L$

Where

$\triangle H_f$ is enthalpy of formation

S is enthalpy of sublimation

B is bond enthalpy

$IE_M$ is ionisation enthalpy of metal

$EA_X$ is electron affinity of non metal atom

$U_L$ is lattice energy

Substituting the given values we have

-435.7 = 79.2 + 1/2 x 242.8 + 418.7 - 348 +U_L

$U_L$ = - 707 KJ / mol

3 0

1 year ago

Lead is malleable, so it can be pounded into flat sheets without breaking. How does the bonding within lead help to explain this

jarptica [38.1K]

The answer is Metallic bonds involve many valence electrons shared by many atoms, so the bonds can move around as the metal is pounded. The metallic bond structure of lead forms a cubic crystal structure and the atoms can roll over one another without breaking the metallic bonds. This is especially because the p orbital electrons of lead can be delocalized and the electrons can be shared with other lead ions in the cubic structure of lead.

9 0

2 years ago

Read 2 more answers

Uranium–232 has a half–life of 68.9 years. A sample from 206.7 years ago contains 1.40 g of uranium–232. How much uranium was or

givi [52]

<u>Answer:</u> The initial amount of Uranium-232 present is 11.3 grams.

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

We are given:

$t_{1/2}=68.9yrs$

Putting values in above equation, we get:

$k=\frac{0.693}{68.9}=0.0101yr^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.0101yr^{-1}$

t = time taken for decay process = 206.7 yrs

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 1.40 g

Putting values in above equation, we get:

$0.0101yr^{-1}=\frac{2.303}{206.7yrs}\log\frac{[A_o]}{1.40}$

$[A_o]=11.3g$

Hence, the initial amount of Uranium-232 present is 11.3 grams.

4 0

1 year ago

Read 2 more answers