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The mass of 0.5 moles of methane is...

The mass of 0.5 moles of methane is

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To find the mass of 0.5 moles of methane (CH₄), we can follow these steps: ### Step 1: Identify the molecular formula of methane. Methane is represented by the chemical formula CH₄. ### Step 2: Calculate the molecular mass of methane. To calculate the molecular mass, we need to add the atomic masses of its constituent elements: - Carbon (C) has an atomic mass of approximately 12 g/mol. - Hydrogen (H) has an atomic mass of approximately 1 g/mol. Since there are 4 hydrogen atoms in methane, we multiply the atomic mass of hydrogen by 4. So, the calculation is: - Molecular mass of CH₄ = (1 × atomic mass of C) + (4 × atomic mass of H) - Molecular mass of CH₄ = (1 × 12 g/mol) + (4 × 1 g/mol) = 12 g/mol + 4 g/mol = 16 g/mol. ### Step 3: Determine the mass of 0.5 moles of methane. We know that 1 mole of methane has a mass of 16 grams. To find the mass of 0.5 moles, we can use the following calculation: - Mass of 0.5 moles of CH₄ = 0.5 moles × mass of 1 mole of CH₄ - Mass of 0.5 moles of CH₄ = 0.5 × 16 g = 8 g. ### Conclusion: The mass of 0.5 moles of methane is 8 grams. ---

To find the mass of 0.5 moles of methane (CH₄), we can follow these steps: ### Step 1: Identify the molecular formula of methane. Methane is represented by the chemical formula CH₄. ### Step 2: Calculate the molecular mass of methane. To calculate the molecular mass, we need to add the atomic masses of its constituent elements: - Carbon (C) has an atomic mass of approximately 12 g/mol. ...
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Knowledge Check

  • The formula of an acid is HXO_(2) . The mass of 0.0242 moles of the acid is 1.657 g. What is the atomic mass of X?

    A
    35.5
    B
    28.1
    C
    128
    D
    19.0
  • If the mass of 0.25 moles of an element Xis 2.25g, the mass of one atom of X is about:

    A
    `1.5 xx 10^(-24)`g
    B
    `2.5 xx 10^(-23)`g
    C
    `1.5 xx 10^(-23)`g
    D
    `2.5 xx 10^(-24)`g
  • The mass of 1 mole of electrons is :

    A
    `9.1 xx 10^(-28) g`
    B
    1.008 mg
    C
    0.55 mg
    D
    `9.1 xx 10^(-27) g`
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