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RSB [31]
1 year ago
5

In addition to reacting with gold (see Problem 21.21), aqua regia is used to bring other precious metals into solution. Balance

the skeleton equation for the reaction with Pt:Pt(s) + NO₃⁻(aq) + Cl⁻(aq) → PtCl₆²⁻(aq) + NO(g)
Chemistry
1 answer:
suter [353]1 year ago
6 0

Aqua regia is used to bring precious metals into the solution .

3Pt(s)+ 4NO^{-} _{3} (aq)+18 Cl^{-} (aq) +16H^{+} =3Pt Cl^{2-} _{6} (aq) +4NO(g) +8 H_{2} O

This is the balanced reaction of Pt (platinum) .

<h3>What is aqua regia ?</h3>

It is a mixture of concentrated nitric acid and hydrochloric acid .there is three part of concentrated HCl and 1 part of concentrated nitric acid .

it is red or yellowish liquid . it is extremely corrosive and can cause skin burns . it is frequently used to dissolve gold and platinum .

The equation of reaction of aqua regia with gold :

Au + 4H^{+} +NO^{-} _{3} +4Cl^{-} =AuCl^{-} _{4} +NO +2H_{2} O

Learn more about aqua regia here :

brainly.com/question/12982022

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A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.692 kg of water
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Answer:

494.49 g of NaCl.

Explanation:

Data obtained from the question include the following:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mass of NaCl =.?

Next, we shall determine the number of mole of NaCl in the solution.

Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as

Molality = mole of solute /Kg of solvent

With the above formula, we can obtain the number of mole NaCl in the solution as follow:

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Mole of NaCl =..?

Molality = mole of solute /Kg of solvent

3.140 = mole of NaCl /2.692

Cross multiply

Mole of NaCl = 3.140 x 2.692

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8.45288 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 8.45288 × 58.5

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Therefore, 494.49 g of NaCl are present in the solution.

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Magnesium reacts steadily with hydrochloric acid
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I don't know what you mean. I can show the chemical equation though.

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