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2 years ago

According to the quantum-mechanical model, which element will have similar chemical properties to an element with z = 7?

1 answer:

7 0

According to the quantum-mechanical model, Phosphorous will have have similar chemical properties to an element with Z = 7

Quantum mechanics models describe the opportunity of putting electrons inside an atom through describing the main power degree, energy degree, orbital (arbitrary degree), and spin of. The quantum mechanics version is based totally on quantum theory, which states that rely has wave-like properties.

The quantum mechanical model is primarily based on quantum concept, which states that matter nonetheless has wave-related residences. The quantum mechanical model is critical for knowledge chemistry since it describes how electrons arise in atoms and how these electrons determine the chemical and physical homes of an detail.

The maximum particular clocks within the global, atomic clocks, are able to use standards of quantum idea to measure time. They screen the specific radiation frequency needed to make electrons leap among energy tiers.

explanation :

Nitrogen (Z = 7) is in Group 15 element

and will have similar properties to other Group 15 elements, like Phosphorous etc.

Learn more about quantum numbers here:-brainly.com/question/5927165

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Which one of the following is not an agent of chemical weathering?

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2 years ago

g Suppose the formation of nitrosyl chloride proceeds by the following mechanism: step elementary reaction rate constant That is

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3 years ago

An OH group attached to a hydrocarbon is called a _________ group whereas ______________ is a polyatomic ion with a charge of __

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3 years ago

The thermochemical equation for the reaction of carbon dioxide with sulfur dioxide is shown below. How can the reaction be descr

prisoha [69]

Answer: d. has high activation energy

Activation energy is the energy that a system requires to start a certain process. Also, it <u>is the minimum energy necessary for a given chemical reaction to occur</u>. For a reaction to occur between two molecules, they must collide in the correct orientation and have a minimum amount of energy equal to the activation energy.

As the molecules approach, their electron clouds repel, so energy is required for the collision to occur and therefore the reaction. The activation energy comes from the heat of the system, that is, the translational, vibrational energy, etc. of each molecule. However, if this energy is not enough, the reaction will not be spontaneous.

<u>A reaction between two molecules can be favored by supplying energy to the system.</u> In the case raised in the question, <u>energy equal to 1104 kJ is provided to the system to favor the next reaction </u>

CO2 (g) + 2SO2 (g) → CS2 (g) + 3O2 (g)

<u>Since the energy equal to 1104 kJ is included in the reactants, it can be deduced that it is the energy that is provided to the system for the reaction to occur. </u>However, from the value of this energy it can not be said whether the system is endothermic or exothermic since it is a kinetic variable and the variables of this type do not allow predicting the thermodynamic behavior of a system.

Furthermore, it can be seen that the value of this energy is considerably high, therefore the reaction described has a high activation energy.

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3 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

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3 years ago