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Current Question (ID: 8662)

Question:

$\text{Atomic hydrogen or oxy-hydrogen torch is used for cutting and welding purposes because:}$

Options:

1. $\text{Atomic hydrogen converts into molecular hydrogen and generates a large amount of energy.}$
2. $\text{Atomic hydrogen converts into molecular hydrogen and generates a low amount of energy.}$
3. $\text{Molecular hydrogen converts into atomic hydrogen and generates a large amount of energy.}$
4. $\text{Molecular hydrogen converts into atomic hydrogen and generates a low amount of energy.}$

Solution:

$\text{Hint: Hydrogen coming out of the electric arc, they reunite to form molecular hydrogen.}$ $\text{Explanation:}$ $\text{Atomic hydrogen atoms are produced by the dissociation of dihydrogen with the help of an electric arc. As soon as the atoms of hydrogen coming out of the electric arc, they reunite to form molecular hydrogen.}$ $\text{This releases a huge amount of energy (435.88 kJ mol}^{-1}\text{). This energy can be used to generate a temperature of 4000 K, which is ideal for welding and cutting metals.}$ $\text{Hence, atomic hydrogen or oxy-hydrogen torches are used for these purposes. For this reason, atomic hydrogen is allowed to recombine on the surface to be welded to generate the desired temperature.}$

Import JSON File

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
}