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laiz [17]
2 years ago
15

In each pair of statements, choose the example that is more reliable.

Chemistry
2 answers:
TEA [102]2 years ago
3 0

Answer:

"statement 2" for the first pair and "statement 1" for the second pair

Explanation:

BlessedZay2 years ago
0 0

In each pair of statements, choose the example that is more reliable.

Statement 1
The company Herbal Remedies Inc. announced today that its antiobesity skin treatment is safe and effective.


Statement 2
The US Food and Drug Administration (FDA) announced today that cancer treatment X, manufactured by Pharma Inc., is safe and effective.


Statement 1
An independent chemical analysis paid for by local residents determined that wells are contaminated with dioxin at a concentration of 3 ppb.


Statement 2
The chemical company Chemicals Inc. insists that local groundwater has not been affected adversely by a leak in its waste handling system.


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1.33 dm3 of water at 70°C are saturated by 2.25
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Given that 4.50 dm³ of Pb(NO₃)₂ is cooled from 70 °C to 18 °C, the

amount amount of solute that will be deposited is 1,927.413 grams.

<h3>How can the amount of solute deposited be found?</h3>

The volume of water 1.33 dm³ of water 70 °C.

The number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water at 70 °C  = 2.25 moles

At 18 °C, the number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water = 0.53 moles

Therefore;

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C is therefore;

1.33 dm³ contains 2.25 moles.

Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{2.25}{1.33} \times 4.50 \approx \mathbf{7.613 \, moles}

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C ≈ 7.613 moles

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C is therefore;

1.33 dm³ contains 0.53 moles

Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{0.53}{1.33} \times 4.50 \approx \mathbf{1.79 \, moles}

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C ≈ 1.79 moles

The number of moles that precipitate out = The amount of solute deposited

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Amount of solute deposited = 7.613 moles - 1.79 moles = 5.823 moles

The molar mass of Pb(NO₃)₂ = 207 g + 2 × (14 g + 3 × 16 g) = 331 g

The molar mass of Pb(NO₃)₂ = 331 g/mol

The amount of solute deposited = Number of moles × Molar mass

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The amount of solute deposited = 5.823 moles × 331 g/mol =<u> 1,927.413 g </u>

Learn more about saturated solutions here:

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