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Feliz [49]
3 years ago
7

When 10.0 g of lead are heated with 1.6 g of sulfur, 11.6 g of lead sulfide are formed. how many grams of lead sulfide form when

10.0 g of lead are heated with 3.0 g of sulfur?
Chemistry
1 answer:
nekit [7.7K]3 years ago
8 0
<span>11.6 g of lead sulfide. First, get the molar masses of lead and sulfur Lead = 207.2 Sulfur = 32.065 Now determine how many moles of each we have avaiable lead = 10.0 g / 207.2 g/mol = 0.048262548 mol sulfur = 1.6 g / 32.065 g/mol = 0.049898643 = mol This tells me that the what's being produced is PbS instead of PbS2 and that there's a very slight excess of sulfur in the original reaction. So on the 2nd reaction with the same amount of lead and twice the amount of sulfur, there will be an even greater excess of sulfur and that you'll get 11.6 g of lead sulfide.</span>
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Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

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V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

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density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

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Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

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144 = \dfrac{\sqrt{3}}{4}a

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In a unit cell, Volume V = a³

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density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

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From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

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