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2 years ago

An example of a solution that is formed endothermically is the one used in

Chemistry

1 answer:

Makovka662 [10]2 years ago

3 0

Cold Packs

Explanation:

Endothermic reactions are reactions in which heat is absorbed from the surroundings and the container in which the reaction takes place is said to be very cold. In cold packs, ammonium nitrate is used, is dissolved in solution is said to be an example of endothermic reaction, in this the heat energy is absorbed from the surroundings. In this the solute increases the solubility with the rise in temperature, and also used in cold packs.

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. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

Answer: The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

Explanation:

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago

Which of the following is false? a. an apoenzyme can catalyze its reaction without its cofactor b an organic cofactor can be cal

ozzi

Answer: Option (a) is the correct answer.

Explanation:

A protein part of an enzyme is known as an apoenzyme. An apoenzyme combines with a cofactor, it is known as holoenzyme.

Without a cofactor an apoenzyme cannot function as cofactor helps in the formation of an active enzyme system and provides a specific site on enzyme for the substrate.

Whereas a non-protein chemical compound or metal ion that helps in the activity of enzyme as a catalyst is known as a cofactor. A metal ion cofactor can be bound directly to the enzyme or to a coenzyme.

The organic non-protein molecules which bind to the protein molecule to form an active enzyme is known as a coenzyme. Coenzymes are small size molecules which help the enzymes to act as a catalyst.

Therefore, we can conclude that the statement an apoenzyme can catalyze its reaction without its cofactor, is false.

3 0

3 years ago

Identify the 2 numbers given for each element

Komok [63]

where are the element you haven't showed it so that we can answer it

6 0

2 years ago

How many moles of Carbon are in 3.06 g of Carbon

natta225 [31]

Answer:

$\boxed {\boxed {\sf 0.255 \ mol \ C }}$

Explanation:

If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

Look up the molar mass of carbon.

Carbon (C): 12.011 g/mol

Set up a ratio using the molar mass.

$\frac {12.011 \ g \ C}{ 1 \ mol \ C}$

Since we are converting 3.06 grams to moles, we multiply by that value.

$3.06 \ g \ C*\frac {12.011 \ g \ C}{ 1 \ mol \ C}$

Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.

$3.06 \ g \ C* \frac{1 \ mol \ C}{12.011 \ g\ C}$

$3.06 * \frac{1 \ mol \ C}{12.011 }$

$\frac {3.06}{12.011 } \ mol \ C$

$0.25476646 \ mol \ C$

The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

0.25476646

The 7 in the ten-thousandth place tells us to round the 4 up to a 5.

$0.255 \ mol \ C$

3.06 grams of carbon is approximately 0.255 moles of carbon.

3 0

2 years ago

Calculate how many times more soluble Mg(OH)2 is in pure water Based on the given value of the Ksp, 5.61×10−11, calculate the ra

maks197457 [2]

Answer:

molar solubility in water = 2.412 * 10^-4 mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

The equation for the solubilization reaction of Mg(OH)2 can be given as:

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

Ksp can then be given as followed:

Ksp = [Mg^2+][OH^–]²

Step 2: Calculate the solubility in water

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X of OH-

The concentration at equilibrium will be XM Mg^2+ and 2X OH-

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X * (2X)² = X *4X² = 4X³

X = 2.412 * 10^-4 mol/L = solubility in water

Step 3: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2+ = 0 M

The initial concentration of OH- = 0.130 M

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X +0.130 for OH-

The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-

The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X *(0.130)²

5.61*10^-11 = X * (0.130)^2

X = 3.32*10^-9 = solubility in 0.130 M NaOH

Step 4: Calculate how many times Mg(OH)2 is better soluble in pure water.

(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

4 0

3 years ago