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Current Question (ID: 7457)

Question:

$\text{The complete symbol for the atom with the given atomic number (Z) and atomic mass (A) would be, respectively:}$ $\text{i. Z = 17, A = 35}$ $\text{ii. Z = 92, A = 233}$ $\text{iii. Z = 4, A = 9}$

Options:

1. ${}^9\text{Be}_4 \quad {}^{233}\text{U}_{92} \quad {}^9\text{Be}_4$
2. ${}^{35}\text{Cl}_{17} \quad {}^{233}\text{U}_{92} \quad {}^9\text{Be}_4$
3. ${}^9\text{Be}_4 \quad {}^{35}\text{Cl}_{17} \quad {}^{233}\text{U}_{92}$
4. ${}^{233}\text{U}_{92} \quad {}^9\text{Be}_4 \quad {}^{35}\text{Cl}_{17}$

Solution:

$\text{HINT: Z = Atomic number; A = atomic mass}$ $\text{To write the complete atomic symbol, we use the format } {}^A\text{X}_Z \text{ where X is the element symbol.}$ $\text{For each case:}$ $\text{i. Z = 17, A = 35:}$ $\text{Atomic number 17 corresponds to Chlorine (Cl)}$ $\text{Complete symbol: } {}^{35}\text{Cl}_{17}$ $\text{ii. Z = 92, A = 233:}$ $\text{Atomic number 92 corresponds to Uranium (U)}$ $\text{Complete symbol: } {}^{233}\text{U}_{92}$ $\text{iii. Z = 4, A = 9:}$ $\text{Atomic number 4 corresponds to Beryllium (Be)}$ $\text{Complete symbol: } {}^9\text{Be}_4$ $\text{Therefore, the correct sequence is: } {}^{35}\text{Cl}_{17}, {}^{233}\text{U}_{92}, {}^9\text{Be}_4$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}