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Rudiy27
2 years ago
14

Suppose you have a solid that looks like gold but you believe it to be fool’s gold. The mass of the solid is 23.5 grams. When th

e solid is lowered into a graduated cylinder the water level rises from 47.5 to 52.2 mL. Is the simple fool’s gold
Chemistry
1 answer:
frutty [35]2 years ago
3 0

Answer:

The sample is fool's gold

Explanation:

Density is defined as the ratio between mass in grams and volume in mililiters.

A sample of pure gold has a density of 19.3g/mL.

Using Archimedes' principle, the volume of the sample is:

52.2mL - 47.5mL = 4.7mL

As the mass of the sample is 23.5g, the density is:

23.5g / 4.7mL = 5g/mL

The denisty of the sample is very different to density of pure gold, that means:

<em>the sample is fool's gold</em>

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Write the formula of the conjugate acid of HCO₂⁻.
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Taking into account the Brønsted-Lowry acid-base theory, the conjugate acid of HCO₂⁻ is H₂CO₂.

<h3>Brønsted-Lowry acid-base </h3>

The Brønsted-Lowry acid-base theory (or the Brønsted-Lowry theory) identifies acids and bases based on whether the species accepts or donates protons or H⁺.

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Then, a conjugate base is an ion or molecule resulting from the acid that loses the proton, while a conjugate acid is an ion or molecule resulting from the base that gains the proton:

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<h3>Conjugate acid of HCO₂⁻</h3>

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end point comes at = 10 ml

So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M

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K_{sp} = [0.01 × (0.02)²] = 4×10⁻⁶

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