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3 years ago

Which of the following is an example of a transuranium element

Chemistry

2 answers:

ella [17]3 years ago

8 0

Elements with atomic number greater than 92(Atomic number of Uranium) are known as transuranium element.

Cm - Curium is a transuranium element as its atomic number is 96.

Murrr4er [49]3 years ago

6 0

$\boxed{{\text{C) Curium, Cm}}}$ is an example of a transuranium element.

Further Explanation:

A tabular chart of the chemical elements in horizontal rows and vertical groups is termed as a periodic table. In this table, elements are arranged in the increasing order of their atomic numbers. The elements are arranged in such a way that those having similar properties are placed together. The creation of the periodic table has made the study of elements simpler and easier.

Transuranium or transuranic elements are the elements that have atomic masses greater than 92. All such elements are highly unstable and undergo radioactive decay. The element with atomic number 92 is uranium. So all the elements that come after uranium in the periodic table are known as transuranium elements. Since these elements show radioactivity, these are used in nuclear reactors and nuclear weapons in order to generate a large amount of energy.

The atomic number of samarium (Sm) is 62.

The atomic number of uranium (U) is 92.

The atomic number of curium (Cm) is 96.

The atomic number of thorium (Th) is 90.

Since the atomic number of curium (96) is greater than that of uranium (92), it is an example of transuranium elements. Therefore option C is the correct answer.

Learn more:

Which property do metalloids share with nonmetals? brainly.com/question/802690
Which properties are characteristic of group 2 elements? brainly.com/question/8226316

Answer details:

Grade: High School

Chapter: Periodic classification of elements

Subject: Chemistry

Keywords: periodic table, elements, atomic numbers, horizontal rows, vertical groups, chemical elements, tabular chart, transuranium element, curium, samarium, uranium, thorium, Cm, Sm, U, Th, 92, 96, 90, 62.

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3 years ago

Calculate the activity coefficients for the following conditions:

uysha [10]

<u>Answer:</u>

<u>For a:</u> The activity coefficient of copper ions is 0.676

<u>For b:</u> The activity coefficient of potassium ions is 0.851

<u>For c:</u> The activity coefficient of potassium ions is 0.794

<u>Explanation:</u>

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

$-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}$ ........(1)

where,

$\gamma_i$ = activity coefficient of ion

$Z_i$ = charge of the ion

$\mu$ = ionic strength of solution

$\alpha _i$ = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

$\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)$ ......(2)

where,

$C_i$ = concentration of i-th ions

$Z_i$ = charge of i-th ions

<u>For a:</u>

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

$C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M$

Now, calculating the activity coefficient of $Cu^{2+}$ ion in the solution by using equation 1:

$Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{ (known)}\\\mu=0.01M$

Putting values in equation 1, we get:

$-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676$

Hence, the activity coefficient of copper ions is 0.676

<u>For b:</u>

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

$C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.025M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851$

Hence, the activity coefficient of potassium ions is 0.851

<u>For c:</u>

We are given:

0.02 M $K_2SO_4$ solution:

Calculating the ionic strength by using equation 2:

$C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.06M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794$

Hence, the activity coefficient of potassium ions is 0.794

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3 years ago