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melomori [17]
3 years ago
9

Determine the value of the equilibrium constant (report your answer to three significant figures) for the following reaction if

an equilibrium mixture contains 0.010 mol of solid PbBr2, and is 0.0100 M in Pb2+ ions and 0.0250 M in Br1- ions. Use the notation 4.31e-5 to indicate a number such as 4.31 x 10-5.
Chemistry
1 answer:
Debora [2.8K]3 years ago
5 0

Answer:

6.25e-6 is the value of the equilibrium constant

Explanation:

we have this equation

PbBr(s) ----- Pb^{2+}(aq)  + 2Br(aq)

When at a state of equilibrium,

we have the concentration of Pb^2+ to be 0.01

we have the concentration of Br^- to be 0.025

the equilibrium constant concentration of both pure solids and liquid s are said to be equal to 1

[PbBR2] = 1

such tht

Keq = [Pb^2+] x [Br-]^2

we already know the values of these from the above.

0.01x0.025^2

= 0.01 x 0.000625

= 0.00000625

= 6.25 x 10^-6

= 6.25e^-6

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Dima020 [189]

Nitrogen (N2) and hydrogen (H2) gases react to form ammonia, which requires -99.4 J/K of standard entropy (ΔS°).

What is standard entropy?

The difference between the total standard entropies of the reaction mixture and the summation of the standard entropies of the outputs is the standard entropy change. Each entropy in the balanced equation needs to be compounded by its coefficient, as shown by the letter "n."

Calculation:

Balancing the given reaction following-

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ΔS° = [1 mol x S° (NH₃)g] - [1/2 mol x S° (N₂)g] - [3/2 mol x S°(H₂)g]

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Insert into the aforementioned equation all the typical entropy values found in the literature:

ΔS° = [1 mol x 192.45 J/mol.K] - [1/2 mol x 191.61 J/mol.K] - [3/2 mol x 130.684 J/mol.K]

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5 0
1 year ago
All organisms need to obtain _?__ from food to survive
PSYCHO15rus [73]
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3 0
3 years ago
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ivann1987 [24]

Answer:

C.

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3p contains (3 electrons) should fill the 3 orbitals firstly. Every orbital contains 1 electron and be in the same spin direction.

So, the right choice is c.

A is wrong because 2 electrons of 3p are paired in the first orbital before filling every orbital.

B is wrong because the 2 electrons of 1s and 2s are in the same direction and also 2 electrons of 3p are paired in the first orbital before filling every orbital.

D is also wrong the 2 electrons of 1s and 2s are in the same direction and the electron in the second orbital of 3p are in opposite direction of the other 2 electrons.

7 0
4 years ago
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