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Allisa [31]
2 years ago
14

Write the expression for the equilibrium constant Kp for the following reaction. (Enclose pressures in parentheses and do NOT wr

ite the chemical formula as a subscript. For example, enter (PNH3)2 as (P NH3)2. If either the numerator or denominator is 1, please enter 1.) PCl5(g) ↔ PCl3(g) Cl2(g)
Chemistry
1 answer:
Liono4ka [1.6K]2 years ago
6 0

Answer and Explanation:

For the following balanced reaction:

PCl₅(g) ↔ PCl₃(g) + Cl₂(g)

We can see that all reactants and products are gases, so it is an homogeneous equilibrium. The expression for the equilibrium constant Kp can be written from the partial pressures (P) of reactants and products as follows:

Kp=\frac{(P PCl_{3})(P Cl_{2})}{(P PCl_{5})}

Where PPCl₃ is the partial pressure of PCl₃ (reactant), PCl₂ is the partial pressure of Cl₂ (reactant) and PPCl₅ is the partial pressure of PCl₅ (product).

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2 years ago
A sample of ice is heated continuously until it becomes a liquid, and then a gas. Its temperature is recorded throughout and a g
padilas [110]

Answer:

The answer to your question is: C. The specific latent heat of fusion

Explanation:

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D. The internal energy measures the energy of the molecules of a substance, so this answer is wrong.

3 0
2 years ago
What is the oxidation number for each atom in NH4CI?
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</span><span>

</span>
5 0
3 years ago
Define the term inertia
n200080 [17]

Answer:

Explanation:

Enertia is an integral part of Newton's first law of motion.

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5 0
3 years ago
An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If t
topjm [15]

Answer:

\boxed{3}

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}

Data:

n_{f} = 2

λ = 657 nm

Calculation:  

\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}

7 0
2 years ago
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