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rosijanka [135]
2 years ago
8

A student used 1.2 g of hydrate in the first trial and 1.6 g in the second trial. Explain fully why her mass percent of water wi

ll be the same for both trials even though she used more hydrate in the second trial. (Do not simply state that the mass percent is an intensive property ).
Chemistry
1 answer:
Pachacha [2.7K]2 years ago
8 0

Both trials of 1.2 g and 1.6 g will have the same mass percent of water because the ratio of the salt to the water of hydration is always constant for any hydrated salt.

<h3>Water of hydration</h3>

For every hydrated salt, the ratio of the salt to the water of hydration remains constant irrespective of the amount of salt taken for experimental analysis.

For example, assuming the mass percent of water in 10g of a hydrated salt is 40%, if 100g of the same salt is taken, the mass percent will remain 40%.

More on water of hydration can be found here: brainly.com/question/11202174

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4 0
3 years ago
the quantity of antimony in an ore can be determined by an oxidation-reduction titration with an oxidizing agent. The ore is dis
Basile [38]

Answer:

BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)

Explanation:

At a redox equation, one substance is being oxidized (losing electrons), and the other is being reduced (gaining electrons). In the given reaction:

BrO₃⁻(aq) + Sb³⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq)

When it's at an acidic solution, it must be ions H⁺ on the reactant, which will form water with the oxygen, so the complete reaction is:

BrO₃⁻(aq) + Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq) + H₂O(l)

As we can see, the antimony is being oxidized (go from +3 to +5), and the Bromo is being reduced. The oxidation number of brome in the reactant, knowing that the oxidation number of O is -2, is:

x + 3*(-2) = -1

x = +5

So, it's going from +5 to -1, and the half-reactions are:

BrO₃⁻(aq) + 6e⁻ → Br⁻(aq)

Sb³⁺(aq) → Sb⁵⁺(aq) + 2e⁻

The number of electrons must be the same, so the second equation must be multiplied by 3:

3Sb³⁺(aq) → 3Sb⁵⁺(aq) + 6e⁻

Thus, the equation will be:

BrO₃⁻(aq) + 3Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + H₂O(l)

Now, we verify the amount of the elements, which must be equal on both sides. So, we multiply H₂O by 3, and H⁺ by 6, and the balanced reaction will be:

BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)

6 0
3 years ago
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Inessa05 [86]

Answer:

True

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mercury metal is poured into a graduated cylinder that holds exactly 22.5 mL. the mercury used to fill the cylinder weighs 306.0
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