1) How many Joules of energy are released when 75g of water is heated

The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−

barxatty [35]

Answer:

The reaction will be non spontaneous at these concentrations.

Explanation:

$AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)$

Expression for an equilibrium constant $K_c$ :

$K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]$

Solubility product of the reaction:

$K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}$

Reaction between Gibb's free energy and equilibrium constant if given as:

$\Delta G^o=-2.303\times R\times T\times \log K_c$

$\Delta G^o=-2.303\times R\times T\times \log K_{sp}$

$\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]$

$\Delta G^o=69,117.84 J/mol=69.117 kJ/mol$

Gibb's free energy when concentration $[Ag^+] = 1.0\times 10^{-2} M$ and $[Br^-] = 1.0\times 10^{-3} M$

Reaction quotient of an equilibrium = Q

$Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}$

$\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)$

$\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])$

$\Delta G=40.588 kJ/mol$

For reaction to spontaneous reaction: $\Delta G$ .
For reaction to non spontaneous reaction: $\Delta G>0$ .

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0

3 years ago

A mixture of so2(g) and o2(g), formed by the complete decomposition of so3(g), is collected over water at 34°c at a total pressu

Sedbober [7]

The decomposition of so3 to so2 gas and o2 gas can be described in the balanced chemical equation:

2so3(g) = 2 so2 (g) + 02(g).

so assuming a complete reaction, the ratio of so2 gas to total products is 2/3 while that of 02 is 1/3.

Subtracting water's water vapor pressure, 760-40 mm hG = 720 mm Hg.

then the products partial pressures are

so2 = 2/3 * (720) = 480 mm Hg.
o2 = 720-480 = 240 mm Hg.

8 0

3 years ago

What is paper chromatography? Calculate the Rr value of a colored dye that traveled 52 mm on a chromatography strip while the so

Karo-lina-s [1.5K]

Answer:

Paper chromatography is a basic technique of chromatography. It consist in the separation of the mixe components using a solvent.

Explanation:

Paper chromatography is a basic technique of chromatography. It consists in the separation of the mixed components using a solvent.

Paper chromatography consists of put some dot of the mix using a glass capillary into a specialized paper, generally made of cellulose, this is called a stationary phase.

Then you put this paper into a camera of glass named, chromatography camera, where previously contain a solvent. The solvent also know as a mobile phase, the type can be defined before the test and involves a study of the kind of the mix, and the compound you want to separate.

The chromatography camera has to be closed all the time during the test, and you can't move at all because the movement of the solvent can alternate the result.

Very often, the solution of the solvent is a mix of different liquid substances with different polarities.

When the stationary phase put into the camera, the solvent starts to move up over the paper, until the separation of the compounds is observable.

the Rf is a value who relates the move of the mobile phase with the move of the distance traveled by the substance tested.

To undersant the paper chromatography, you can watch the images attached.

The first is an image of the chromatography camera.

The second one is an image of a cellulose paper after the chromatography is done. You can watch the dots who indicates the traveling of the compound across the paper.

The third one can show you the evolutions of paper chromatography, from the beginning to the end.

To calculate the Rf value you have to use the equation:

Rf = distance traveled by the substance/distance traveled by the solvent/

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3 0

3 years ago

Determine the percent ionization of a 0.225 M solution of benzoic acid. Express your answer using two significant figures.

Assoli18 [71]

Answer:

1.68% is ionized

Explanation:

The Ka of benzoic acid, C₇H₆O₂, is 6.46x10⁻⁵, the equilibrium in water of this acid is:

C₇H₆O₂(aq) + H₂O(l) ⇄ C₇H₅O₂⁻(aq) + H₃O⁺(aq)

Ka = 6.46x10⁻⁵ = [C₇H₅O₂⁻] [H₃O⁺] / [C₇H₆O₂]

<em>Where [] are concentrations in equilibrium</em>

In equilibrium, some 0.225M of the acid will react producing both C₇H₅O₂⁻ and H₃O⁺, the equilibrium concentrations are:

[C₇H₆O₂] = 0.225-X

[C₇H₅O₂⁻] = X

[H₃O⁺] = X

Replacing:

6.46x10⁻⁵ = [X] [X] / [0.225-X]

1.4535x10⁻⁵ - 6.46x10⁻⁵X = X²

1.4535x10⁻⁵ - 6.46x10⁻⁵X - X² = 0

Solving for X:

X = -0.0038. False solution, there is no negative concentrations.

X = 0.00378M. Right solution.

That means percent ionization (100 times Amount of benzoic acid ionized over the initial concentration of the acid) is:

0.00378M / 0.225M * 100 =

<h3>1.68% is ionized</h3>

8 0

3 years ago

How many liters of 15.0 molar NaOH stock solution will be needed to make 17.5 liters of a 1.4 molar NaOH solution? Show the work

strojnjashka [21]

2.0 L
The key to any dilution calculation is the dilution factor

The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.

In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to

DF=18.5M1.5M=12.333

So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.

The volume of the stock solution needed for this dilution will be

DF=VdilutedVstock⇒Vstock=VdilutedDF

Plug in your values to find

Vstock=25.0 L12.333=2.0 L−−−−−

The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.

So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.

IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!

In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.

Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!

So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.

Always remember

Water to concentrated acid →.NO!

Concentrated acid to water →.YES!

8 0

3 years ago