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Elenna [48]
2 years ago
14

Identify the correct equation for the equilibrium constant Kc for the reaction given. C u ( s ) + 2 A g N O 3 ( a q ) − ⇀ ↽ − C

u ( N O 3 ) 2 ( a q ) + 2 A g ( s ) Cu(s)+2AgNOX3(aq)↽−−⇀Cu(NOX3)X2(aq)+2Ag(s) Select one: K c = [ Cu ( NO 3 ) 2 ] [ Ag ] 2 [ AgNO 3 ] 2 [ Cu ] Kc=[Cu(NO3)2][Ag]2[AgNO3]2[Cu] K c = [ Cu ( NO 3 ) 2 ] 2 [ Ag ] [ AgNO 3 ] [ Cu ] 2 Kc=[Cu(NO3)2]2[Ag][AgNO3][Cu]2 K c = [ Cu ( NO 3 ) 2 ] [ AgNO 3 ] 2 Kc=[Cu(NO3)2][AgNO3]2 K c = [ Cu ( NO 3 ) 2 ] [ AgNO 3 ] Kc=[Cu(NO3)2][AgNO3]
Chemistry
1 answer:
lana [24]2 years ago
3 0

Answer:

Kc = [Cu(NO₃)₂]/[2 AgNO₃]²

Explanation:

Let's consider the following balanced redox equation.

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

The concentration equilibrium constant (Kc) is equal to the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.

The concentration equilibrium constant for this reaction is:

Kc = [Cu(NO₃)₂]/[2 AgNO₃]²

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A well-insulated, closed device claims to be able to compress 100 mol of propylene, acting as a SoaveRedlich-Kwong gas and with
Setler79 [48]

Explanation:

The given data is as follows.

    Moles of propylene = 100 moles,    T_{i} = 300 K

    T_{f} = 800 K,    V_{i} = 2 m^{3}

    V_{f} = 0.02 m^{3},   C_{p} of propylene = 100 J/mol

Now, we assume the following assumptions:

Since, it is a compression process therefore, work will be done on the system. And, work done will be equal to the heat energy liberating without any friction.

            W = mC_{p} \Delta T

     100 moles \times 100 J/mol K (800 - 300) K

                 = 5 \times 10^{6} J

                 = 5 MJ

Thus, we can conclude that a minimum of 5 MJ work is required without any friction.

3 0
2 years ago
How many atoms of zirconium are in 0.3521 mol of zirconium?
lora16 [44]

Answer:

2.12×10²³ atoms.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This simply means that 1 mole of zirconium also 6.02×10²³ atoms.

Thus, we can obtain the number of atoms present in 0.3521 mole of zirconium as follow:

1 mole of zirconium also 6.02×10²³ atoms.

Therefore, 0.3521 mole of zirconium will contain = 0.3521 × 6.02×10²³ = 2.12×10²³ atoms.

Therefore, 0.3521 mole of zirconium contains 2.12×10²³ atoms.

3 0
2 years ago
A 15.0 mL sample of 0.013 M HNO3 is titrated with 0.017 M CH$NH2 which he Kb=3.9 X 10-10. Determine the pH at these points: At t
kramer

<u>Answer:</u> The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is

<u>Explanation:</u>

  • <u>For 1:</u> At the beginning

To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

Nitric acid is a monoprotic acid and it dissociates 1 mole of hydrogen ions. So, the concentration of hydrogen ions is 0.013 M

[H^+]=0.013M

Putting values in above equation, we get:

pH=-\log(0.013)\\\\pH=1.89

  • <u>For 2:</u>

To calculate the number of moles, we use the equation:  

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

  • <u>For nitric acid:</u>

Molarity of nitric acid solution = 0.013 M

Volume of solution = 15 mL

Putting values in above equation, we get:

0.013M=\frac{\text{Moles of }HNO_3\times 1000}{15}\\\\\text{Moles of }HNO_3=1.95\times 10^{-4}mol

  • <u>For methylamine:</u>

Molarity of methylamine solution = 0.017 M

Volume of solution = 10 mL

Putting values in above equation, we get:

0.017M=\frac{\text{Moles of }CH_3NH_2\times 1000}{10}\\\\\text{Moles of }CH_3NH_2=1.7\times 10^{-4}mol

  • The chemical equation for the reaction of nitric acid and methylamine follows:

                       HNO_3+CH_3NH_2\rightarrow CH_3NH_3^++NO_3^-

As, the mole ratio of nitric acid and methyl amine is 1 : 1. So, the limiting reagent will be the reactant whose number of moles are less, which is methyl amine.

By Stoichiometry of the reaction:

1 mole of methyl amine produces 1 mole of CH_3NH_3^+

So, 1.7\times 10^{-4}mol of methyl amine will produce = \frac{1}{1}\times 1.7\times 10^{-4}=1.7\times 10^{-4}\text{ moles of }CH_3NH_3^+

To calculate the pK_b of base, we use the equation:

pK_b=-\log(K_b)

where,

K_b = base dissociation constant = 3.9\times 10^{-10}

Putting values in above equation, we get:

pK_b=-\log(3.9\time 10^{-10})\\\\pK_b=9.41

  • To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

pOH=pK_b+\log(\frac{[salt]}{[base]})

pOH=pK_b+\log(\frac{[CH_3NH_3^+]}{[CH_3NH_2]})

We are given:

pK_b=9.41

[CH_3NH_3^+]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M

[CH_3NH_2]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M

Putting values in above equation, we get:

pOH=9.41+\log(\frac{6.8\times 10^{-6}}{6.8\times 10^{-6}})\\\\pOH=9.41

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH=14-9.41=4.59

Hence, the pH of the solution is 4.59

4 0
2 years ago
If a typical antacid tablet contains 2.0 g of sodium hydrogen carbonate, how many moles of carbon dioxide should one tablet yiel
ivolga24 [154]
The equation of the chemical reaction is NaHCO3 + H+ --> H2O + CO2 + Na
To determine the total number of moles of carbon dioxide, the given mass of sodium hydrogen carbonate is divided with its own molar mass. Then it is multiplied with the ratio between NaHCO3 and carbon dioxide. The total number of moles of CO2 one tablet should yield is 0.024 mole.
6 0
2 years ago
Read 2 more answers
One of the emission spectral lines for Be31 has a wavelength of 253.4 nm for an electronic transition that begins in the state w
max2010maxim [7]

Explanation:

The given data is as follows.

   \lambda = 253.4 nm = 253.4 \times 10^{-9}m      (as 1 nm = 10^{-9})

            n_{1} = 5,        n_{2} = ?

Relation between energy and wavelength is as follows.

                    E = \frac{hc}{\lambda}

                       = \frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{253.4 \times 10^{-9}}

                       = 0.0784 \times 10^{-17} J

                       = 7.84 \times 10^{-19} J

Hence, energy released is 7.84 \times 10^{-19} J.

Also, we known that change in energy will be as follows.

     \Delta E = -2.178 \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}

where, Z = atomic number of the given element

 7.84 \times 10^{-19} J = -2.178 \times (4)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}

    \frac{7.84 \times 10^{-19} J}{34.848} = \frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}

      0.02 + 0.04 = \frac{1}{n^{2}_{1}}

                      n_{1} = \sqrt{\frac{1}{0.06}}

                          = 4

Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.

4 0
2 years ago
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