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Which of the following reactions would have the smallest value of K at 298 K? Which of the following reactions would have the smallest value of K at 298 K? A + B → 2 C; E°cell = -0.030 V A + 2 B → C; E°cell = +0.98 V A + B → C; E°cell = +1.22 V A + B → 3 C; E°cell = +0.15 V More information is needed to determine.

Answers

Answer 1

Answer:

The reactions that would have the smallest value of K is

A + B → 2 C; E°cell = -0.030 V

Option A

Generally the equation for the number of electrons transferred is mathematically given as

$nFE^o_(cell)=RT\ln K$

where

T= Temperature

F=25C(298K)

R = Gas constant

R= 8.314 J/K.mol

F = Faraday's constant

F= 96500 C

We see from the equation that the E-cell is directly proportional to K(equilibrium constant of the reaction)

Hence, The reactions that would have the smallest value of K is

A + B → 2 C; E°cell = -0.030 V

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Answer 2

Answer:

Answer:

The reaction with smallest value of K is :

A + B → 2 C; E°cell = -0.030 V

Explanation:

$nFE^o_(cell)=RT\ln K$

where :

n = number of electrons transferred

F = Faraday's constant = 96500 C

$E^o_(cell)$ = standard electrode potential of the cell

R = Gas constant = 8.314 J/K.mol

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

$K$ = equilibrium constant of the reaction

As we cans see, that standard electrode potential of the cell is directly linked to the equilibrium constant of the reaction.

Higher $E^o_(cell)$ higher will be the value of K.
Lower $E^o_(cell)$ lower will be the value of K.

So, the reaction with smallest value of electrode potential will have smallest value of equilibrium constant. And that reaction is:

A + B → 2 C; $E^o_(cell) =-0.030 V$

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Be sure to answer all parts. The first step in HNO3 production is the catalyzed oxidation of NH3. Without a catalyst, a different reaction predominates: 4 NH3(g) +3 O2 (g) ⇌ 2 N2(g) + 6 H2O(g) When 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00−L container at a certain temperature, the N2 concentration at equilibrium is 1.96 × 10−3 M. Calculate Kc.

Answers

Answer: The value of $K_c$ for the reaction is $6.005* 10^(-6)$

Explanation:

We are given:

Initial moles of $NH_3=0.0150mol$

Initial moles of $O_2=0.0150mol$

Volume of the container = 1.00 L

Molarity of the solution = $\frac{\text{Number of moles}}{\text{Volume of container}}$

$[NH_3]_i=(0.0150)/(1.00)=0.0150M$

$[O_2]_i=(0.0150)/(1.00)=0.0150M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)$

Initial: 0.0150 0.0150

At eqllm: 0.0150-4x 0.0150-3x 2x 6x

The expression of $K_c$ for above equation follows:

$K_c=([N_2]^2[H_2O]^6)/([NH_3]^4[O_2]^3)$ .......(1)

We are given:

Equilibrium concentration of $N_2=1.96* 10^(-3)$

Equating the equilibrium concentrations of nitrogen, we get:

$2x=1.96* 10^(-3)\n\nx=0.98* 10^(-3)M$

Calculating the equilibrium concentrations:

Concentration of $NH_3=(0.0150-4x)=0.0150-4(0.00098)=0.01108M$

Concentration of $O_2=(0.0150-3x)=0.0150-3(0.00098)=0.01206M$

Concentration of $N_2=2x=2(0.00098)=0.00196M$

Concentration of $H_2O=6x=6(0.00098)=0.00588M$

Putting values in expression 1, we get:

$K_c=((0.00196)^2* (0.00588)^6)/((0.01108)^4* (0.01206)^3)\n\nK_c=6.005* 10^(-6)$

Hence, the value of $K_c$ for the reaction is $6.005* 10^(-6)$

Final answer:

To calculate the equilibrium constant, Kc, for the reaction that produces HNO3 from NH3 and O2, you need to determine the equilibrium concentrations of NH3 and O2. The given information includes the initial moles and concentration of NH3 and O2, as well as the equilibrium concentration of N2. Using the stoichiometry of the reaction and the given data, you can calculate the equilibrium concentrations and substitute them into the Kc expression to determine the numerical value of Kc.

Explanation:

The question asks to calculate the equilibrium constant, Kc, for the reaction that produces HNO3 from NH3 and O2. The reaction equation is 4 NH3(g) + 3 O2(g) ⇌ 2 N2(g) + 6 H2O(g). The given information is that 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00-L container, and the N2 concentration at equilibrium is 1.96 × 10−3 M. To solve for Kc, we need to calculate the equilibrium concentrations of NH3 and O2.

Using the stoichiometry of the reaction, we can determine that the equilibrium concentration of NH3 is (0.0150 - 2*1.96 × 10−3) M and the equilibrium concentration of O2 is (0.0150 - 3*1.96 × 10−3) M. Substituting these values into the equilibrium expression for Kc, we can calculate the value of Kc.

In this case, the equilibrium constant, Kc, can be calculated as [N2]^2 / ([NH3]^4 * [O2]^3). Substitute the given equilibrium concentration of N2 and the calculated equilibrium concentrations of NH3 and O2 into the Kc expression to determine the numerical value of Kc.

Learn more about Calculating equilibrium constant here:

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Guys i need a long inforation about....
"Formation of colloids"
Please and thank you

Answers

Answer:

Colloids

There are two basic methods of forming a colloid: reduction of larger particles to colloidal size, and condensation of smaller particles (e.g., molecules) into colloidal particles. Some substances (e.g., gelatin or glue) are easily dispersed (in the proper solvent) to form a colloid; this spontaneous dispersion is called peptization. A metal can be dispersed by evaporating it in an electric arc; if the electrodes are immersed in water, colloidal particles of the metal form as the metal vapor cools. A solid (e.g., paint pigment) can be reduced to colloidal particles in a colloid mill, a mechanical device that uses a shearing force to break apart the larger particles. An emulsion is often prepared by homogenization, usually with the addition of an emulsifying agent. The above methods involve breaking down a larger substance into colloidal particles. Condensation of smaller particles to form a colloid usually involves chemical reactions—typically displacement, hydrolysis, or oxidation and reduction.

What is the mass of silver that can be prepared from 1.50 g of copper metal? Cu(s)+2AgNO3(aq)→Cu(NO3)2(aq)+2Ag(s)

Answers

The mass of silver that can be prepared from 1.50 g of copper metal is 5.10 g.

To find the mass of silver that can be prepared from 1.50 g of copper metal, we need to use stoichiometry and the balanced chemical equation.

From the equation, we can see that 1 mole of copper reacts with 2 moles of silver to produce 1 mole of copper(II) nitrate and 2 moles of silver.

We can use the molar masses of copper and silver to convert grams to moles, and then use the mole ratio to find the moles of silver. Finally, we can convert moles of silver back to grams using the molar mass of silver.

Step 1: Convert grams of copper to moles of copper. (1.50 g Cu) / (63.55 g/mol Cu) = 0.0236 mol Cu

Step 2: Use the mole ratio of silver to copper from the chemical equation. (0.0236 mol Cu) × (2 mol Ag / 1 mol Cu) = 0.0473 mol Ag

Step 3: Convert moles of silver to grams of silver. (0.0473 mol Ag) × (107.87 g/mol Ag) = 5.10 g Ag

Molar mass:

Ag = 107.86 g/mol
Cu = 63.54 g/mol

Mole ratio:

Cu(s)+2 AgNO₃(aq)→Cu(NO₃)₂(aq)+2 Ag(s)

63.54 g Cu ------------- 2 x 107.86 g Ag
1.50 g Cu -------------- ??

Mass Ag = 1.50 x 2 x 107.86 / 63.54

Mass Ag = 32358 / 63.54

= 509.25 g of Ag

hope this helps!

Bobby created a dilution of 1/100 of a bacterial sample by adding 1 mL of sample to 99 mL of saline. Unfortunately, after Bobby completed the dilution, he knocked the container over spilling the majority of the diluted sample out. After cleaning up the mess, he found he had 19 mL of diluted sample remaining. Can he still completed the microbial count and if so, then write out the steps on how would he determine the original cell concentration of his total remaining samp

Answers

Answer:

There is no short answer.

Explanation:

In the given example Bobby is creating a solution for his bacteria count which consists of 1% bacterial sample.

Considering that the solution was mixed homogeneously, he can apply the procedure to the remaining sample and get the results he wants.

Or if the average number of bacteria in a 1 mL sample is known, he can apply that information proportionally to the 100 mL mixture and find the original cell concentration.

I hope this answer helps.

The molecule XeF2 does not obey the octet rule. Draw its Lewis structure and state the type of octet-rule exception. Include lone pair electrons in the Lewis structure.

Answers

Answer: See attachment.

Explanation:

Exceptions to the octet rule fall into three categories:

An incomplete octet.
An odd number of electrons.
More than eight valence electrons around the central atom.

In addition to the 3s and 3p orbitals, xenon also has 3d orbitals that can be used in bonding. These orbitals enable xenon to form an expanded octet.

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