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Greeley [361]
3 years ago
7

Compared with the freezing-point depression of a 0.01 m c6h12o6 solution, the freezing-point depression of a 0.01 m nacl solutio

n is
Chemistry
1 answer:
agasfer [191]3 years ago
6 0

Answer:

Twice  as much.

Explanation:

That's because the freezing point depression depends on the total number of solute particles.

C₆H₁₂O₆(s) ⟶ C₆H₁₂O₆(aq)

0.01 mol of C₆H₁₂O₆ gives 0.01 mol of solute particles.

NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)

1 mol of NaCl gives 0.01 mol of Na⁺(aq) and 0.01 mol of Cl⁻(aq).

That's 0.02 mol of particles, so the freezing point depression of 0.01 mol·L⁻¹ NaCl will be twice that of 0.01 mol·L⁻¹ C₆H₁₂O₆.

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Answer:

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Explanation:

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Percent error = Found value - accepted value / accepted value * 100

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Percent error = 2.85 g/cm3 - 2.70 g/cm3 / 2.70 g/cm3 * 100

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7.23 J

Explanation:

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2 years ago
How many atoms of N are in 0.82 g of NaNO3? ( molar conversion)
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The number of atoms N = 5.8 x 10²¹

<h3>Further explanation</h3>

A mole is a unit of many particles (atoms, molecules, ions) where 1 mole is the number of particles contained in a substance that is the same amount as many atoms in 12 gr C-12  

1 mole = 6.02.10²³ particles  

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\tt N=n\times No\\\\N=0.0096\times 6.02\times 10^{23}\\\\N=5.8\times 10^{21}~atoms

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