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2 years ago

Which of the following atoms would be expected to be diamagnetic in the ground state?

Chemistry

1 answer:

Rufina [12.5K]2 years ago

8 0

The atoms that would be expected to be diamagnetic in the ground state is magnesium

The magnetism of an atom refers to its electronic configuration. A diamagnetic atom is an atom whose electrons are all paired.

A paired electron is an electron that occurs in pairs in its orbital shell.

At their respective ground state, the electronic configuration of the given elements are as follows:

The electronic configuration of magnesium is 1s²2s²2p⁶3s². As such its a diamagnetic atom.

The electronic configuration of Potassium is 1s²2s²2p⁶3s²3p⁶4s¹. Hence, Potassium has one unpaired electron in its outermost shell.

The electronic configuration of Chlorine is 1s²2s²2p⁶3s²3p⁵. Hence, Chlorine has one unpaired electron in its outermost shell.

The electronic configuration of Cobalt is 1s²2s²2p⁶3s²3p⁶3d⁷4s². Hence, the unpaired electrons of Cobalt in its outermost shell are three.

Therefore, the atoms that are diamagnetic in the ground state is magnesium.

Learn more about diamagnetic atoms here:

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If 9.8g water is used in electrolysis, what is the percent yield if 5.6g of oxygen was

ANEK [815]

Answer:

Approximately $64\%$ .

Explanation:

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

The actual yield of $\rm O_2$ was given. The theoretical yield needs to be calculated from the quantity of the reactant.

Balance the equation for the hydrolysis of water:

$\rm 2\, H_2O \, (l) \to 2\, H_2\, (g) + O_2\, (g)$ .

Note the ratio between the coefficient of $\rm H_2O\, (g)$ and $\rm O_2\, (g)$ :

$\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ .

This ratio will be useful for finding the theoretical yield of $\rm O_2\, (g)$ .

Look up the relative atomic mass of hydrogen and oxygen on a modern periodic table.

$\rm H$ : $1.008$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm H_2O$ and $\rm O_2$ :

$M(\mathrm{H_2O}) =2\times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{O_2}) =2\times 15.999 = 31.998\; \rm g \cdot mol^{-1}$ .

Calculate the number of moles of molecules in $9.8\; \rm g$ of $\rm H_2O$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{9.8\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \approx 0.543991\;\rm g \cdot mol^{-1}$ .

Make use of the ratio $\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ to find the theoretical yield of $\rm O_2$ (in terms of number of moles of molecules.)

$\begin{aligned} n(\mathrm{O_2}) &= \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} \cdot n(\mathrm{H_2O}) \\ &\approx \frac{1}{2} \times 0.543991\; \rm mol \approx 0.271996\; \rm mol \end{aligned}$ .

Calculate the mass of that approximately $0.271996\; \rm mol$ of $\rm O_2$ (theoretical yield.)

$\begin{aligned}m(\mathrm{O_2}) &= n(\mathrm{O_2}) \cdot M(\mathrm{O_2}) \\ &\approx 0.271996\; \rm mol \times 31.998\; \rm g \cdot mol^{-1} \approx 8.70331 \; \rm g \end{aligned}$ .

That would correspond to the theoretical yield of $\rm O_2$ (in term of the mass of the product.)

Given that the actual yield is $5.6\; \rm g$ , calculate the percentage yield:

$\begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \\ &\approx \frac{5.6\; \rm g}{8.70331\; \rm g} \times 100\% \approx 64\%\end{aligned}$ .

4 0

2 years ago

Place the particles in order from smallest to largest.

yulyashka [42]

Answer:

smallest to largest:

Electron, Neutron, Atom, Molecule, Nucleus

Explanation:

sorry if it's not right

7 0

2 years ago

In the following net ionic equation, identify each reactant as either a Bronsted-Lowry acid or a Bronsted-Lowry base. HCN(aq) H2

vfiekz [6]

Answer:

Explanation:

The definition of acids and bases by Arrhenius Theory was modified and extended by Bronsted-Lowry.

Bronsted-Lowry defined acid as a molecule or ion which donates a proton while a base is a molecule or ions that accepts the proton. This definition can be extended to include acid -base titrations in non-aqueous solutions.

In this theory, the reaction of an acid with a base constitutes a transfer of a proton from the acid to the base.

From the given information:

$\mathsf{HCN _{(aq)} + H_2O_{(l)} \to CN^{-}_{(aq)} + H_3O_{(aq)}}$

From above:

We will see that HCN releases an H⁺ ion, thus it is a Bronsted-Lowry acid

$H_2O$ accepts the H⁺ ion ,thus it is a Bronsted-Lowry base.

The formula of the reactant that acts as a proton donor is <u>HCN</u>

The formula of the reactant that acts as a proton acceptor is <u>H2O</u>

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2 years ago

List the path of the respiratory system

Lena [83]

nasal cavities (or oral cavity) > pharynx > trachea > primary bronchi (right & left) > secondary bronchi > tertiary bronchi > bronchioles > alveoli (site of gas exchange)

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3 years ago

43 moles feci3 are dissolved in 0.64l of solution what is the molarity of the solution

nasty-shy [4]

Molarity=moles/liter
molarity=43/0.64
molarity=67.19moles/litre

4 0

3 years ago