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The density of liquid mercury is 13.6 g/...

The density of liquid mercury is `13.6 g//cm^(3).` How many moles of mercury are there in 1 litre of the metal? (Atomic mass of `Hg=200).`

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To solve the problem, we need to determine how many moles of mercury are present in 1 liter of the metal, given its density and atomic mass. Here’s a step-by-step solution: ### Step 1: Understand the given data - Density of mercury (Hg) = 13.6 g/cm³ - Volume of mercury = 1 liter = 1000 mL - Atomic mass of mercury (Hg) = 200 g/mol ### Step 2: Convert volume from liters to cubic centimeters 1 liter is equivalent to 1000 mL, and since 1 mL is equal to 1 cm³, we have: \[ 1 \text{ L} = 1000 \text{ cm}^3 \] ### Step 3: Calculate the mass of mercury in 1 liter Using the density formula: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \] We can rearrange this to find mass: \[ \text{Mass} = \text{Density} \times \text{Volume} \] Substituting the values: \[ \text{Mass} = 13.6 \text{ g/cm}^3 \times 1000 \text{ cm}^3 = 13600 \text{ g} \] ### Step 4: Calculate the number of moles of mercury Using the formula for moles: \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \] Substituting the values: \[ \text{Moles} = \frac{13600 \text{ g}}{200 \text{ g/mol}} = 68 \text{ moles} \] ### Conclusion The number of moles of mercury in 1 liter of the metal is **68 moles**. ---

To solve the problem, we need to determine how many moles of mercury are present in 1 liter of the metal, given its density and atomic mass. Here’s a step-by-step solution: ### Step 1: Understand the given data - Density of mercury (Hg) = 13.6 g/cm³ - Volume of mercury = 1 liter = 1000 mL - Atomic mass of mercury (Hg) = 200 g/mol ### Step 2: Convert volume from liters to cubic centimeters ...
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Knowledge Check

  • The density of mercury is 13.6 g mL^(-1) . Calculate the approximate diameter of an atom of mercury assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom.

    A
    3.01 Å
    B
    2.54 Å
    C
    0.29 Å
    D
    2.91 Å
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