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Alja [10]
4 years ago
8

For each of the following equilibria, write the equilibrium constant expression for Kc. Where appropriate, also write the equili

brium constant expression for Kp.WO3(s)+3H2(g)?W(s)+3H2O(g)
a. Kc=[H2]3[H2O]3
b. Kc=[H2O]3[W][H2]3[WO3]
c. Kc=[H2O]3[H2]3
d. Kc=[H2]3[WO3][H2O]3[W]
Chemistry
2 answers:
valentina_108 [34]4 years ago
6 0

Answer:

option C is the correct answer

c. Kc=[H2O]³/[H2]³

Explanation:

Kp = WO3(s)+3H2(g)?W(s)+3H2O(g)

equilibrium constant expression

 k = \frac{product}{reactant}

=\frac{ [H_20]^3}{H_2]^3}

Solids and liquids are not included in the equilibrium constant expression. This is because they do not affect the reactant amount at equilibrium in the reaction, so they are disregarded and kept at 1

saw5 [17]4 years ago
3 0

Answer:

Answer in explanation

Explanation:

The question asks to write the equilibrium constant expression for the equation. This can be written by placing each of the products in square brackets and raise them to their number of moles, divided by the reactants each raised too, based on number of moles.

The square brackets represent the concentration of the compounds. We write the equilibrium constant expression as follows:

Kc = [H2O]^3[W]/[H2]^3[WO3]

For the equilibrium constant in terms of pressure, we use only the reactants and products that are gaseous.

Kp = [H2O]^3/[H2]^3

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The energy of any one-electron species in its nth state (n = principal quantum number) is given by E = –BZ2 /n2 where Z is the c
Ivahew [28]

Explanation:

(a) The given data is as follows.

            B = 2.180 \times 10^{-18} J

            Z = 4 for Be

Now, for the first excited state n_{f} = 2; and n_{i} = \infinity if it is ionized.

Therefore, ionization energy will be calculated as follows.

         I.E = \frac{-Bz^{2}}{\infinity^{2}} - (\frac{-2.180 \times 10^{-18} J /times (4)^{2}}{(2)^{2}})

              = 8.72 \times 10^{-18} J

Converting this energy into kJ/mol as follows.

           8.72 \times 10^{-18} J \times 6.02 \times 10^{23} mol  

           = 5249 kJ/mol

Therefore, the ionization energy of the Be^{3+} ion in its first excited state in kilojoules per mole is 5249 kJ/mol.

(b) Change in ionization energy is as follows.

         \Delta E = -Bz^{2}(\frac{1}{(4)^{2}} - {1}{(2)^{2}}) = \frac{hc}{\lambda}

   \frac{hc}{\lambda} = 0.1875 \times 2.180 \times 10^{-18} J \times (4)^{2}                

        \lambda = \frac{6.626 \times 10^{-34} \times 2.998 \times 10^{8} m/s}{0.1875 \times 2.180 \times 10^{-18} J \times 16}

                     = 303.7 \times 10^{-10} m

or,                 = 303.7^{o}A

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5 0
3 years ago
Is the volume of a coffee cup closer to 300 cm cubed or 300 dm cubed
riadik2000 [5.3K]

Answer:

It is closer to 300 cm cubed

Explanation:

The volume of a coffee cup is closer to 300 cm cubed because their american volumes are around 10 ounces, and 10 ounces equals about 300 cm cubed. Hope it helps!

8 0
4 years ago
What is the cell potential for the reaction mg(s+fe2+(aq?mg2+(aq+fe(s at 77 ?c when [fe2+]= 3.40 m and [mg2+]= 0.210 m . express
Galina-37 [17]
First, you need to calculate the standard cell potential using standard reduction potential from a textbook or online. Since Mg becomes Mg+2, magnesium is being oxidized because it is losing electrons, you need to flip its potential

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Mg+2 + 2e- --> Mg                potential= -2.37


Cell potential= (-0.44) + (+2.37)= 1.93 V

Now, you need to use Nernst formula to get the answer. I have attached a PDF with the work.
Download pdf
8 0
3 years ago
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