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elena55 [62]
3 years ago
10

A sample of solid NH 4HS is placed in a closed vessel and allowed to equilibrate. Calculate the equilibrium partial pressure (at

m) of ammonia, assuming that some solid NH 4HS remains.
Chemistry
1 answer:
UNO [17]3 years ago
5 0

Answer:

The answer to the question is

The equilibrium partial pressure (atm) of ammonia, assuming that some solid NH₄HS remains 0.26 atm.

Explanation:

To solve the question, we write out the chemical equation as follows

NH₄HS (s) ⇄ NH₃ (g) + H₂S (g)

From the above equation, it is observed that only the gaseous products contribute to the partial pressure

Kp =PNH₃·PH₂S where at Kp = 0.070 and PNH₃, PH₂S are the partial pressures of the gases

However since the number of moles of both gases are equal, therefore by Avogadro's law PNH₃ = PH₂S

Then PNH₃  = √(0.07) = PH₂S = 0.2645 atm. ≅ 0.26 atm.

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6 0
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The balanced equation below represents the decomposition of potassium chlorate.
dezoksy [38]
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The high concentration of carbon dioxide (CO2) in the atmosphere of Mars inhibits infrared radiation from escaping and keeps the
Zarrin [17]

Answer:

B

Explanation:

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3 0
3 years ago
For the following questions, use a periodic table and your atomic calculations to find the unknown information about each isotop
mafiozo [28]

Answer:

Option B is correct.

4

Explanation:

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All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example, if neutral atom has 6 protons than it must have 6 electrons. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.

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2 + 2 = 4

Thus, option B is correct.

7 0
3 years ago
Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25
o-na [289]
The equation Eºcell = 0.0592/n logK must be used to find n and also Eºcell 
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To balance the equation, 6 moles of electrons must be transferred (2 Al and 3 Mg). This will be the value of n in the equation. 
To find Eºcell, you need the reduction potentials which should be given in a table, and given above. Eºcell = -1.66 - (-2.37) = 0.71 V log K = Eºcell x n/0.0592 = 0.71 x 6/0.0592 log K = 71.95 K = 10^71.95 K = 1.1x10^72
6 0
3 years ago
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