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viva [34]
2 years ago
9

Gold has always been a highly prized metal, and it has been widely used from the beginning of history as a store of value. It do

es not rust like iron and does not become tarnished like silver. It is so chemically inert that it will not react with even the strongest concentrated acids. But it can be dissolved in aqua regia – a fresh-prepared mixture of concentrated HNO3 and HCl (1:3).
When Germany invaded Denmark in World War II, the Hungarian chemist George de Hevesy dissolved the gold Nobel Prizes of Max von Laue and James Franck in aqua regia to prevent the Nazis from stealing them. He placed the jar with the solution on a shelf in his laboratory, and after the war, precipitated the gold out of the acid and returned it to the Royal Swedish Academy of Sciences and the Nobel Foundation who recast the medals and again presented them to Laue and Franck.
The unbalanced equation for the reaction of gold with aqua regia is given below.
Add the stoichiometric coefficients to the equation to balance it.

Au(s) + HNO3(aq) + HCl(aq) → HAUCl4(aq) + NO2(g) + H20(l)

What's the function of HCL?
Chemistry
1 answer:
Greeley [361]2 years ago
8 0

Answer: The balanced equation is

Au(s) + 3HNO3(aq) + 4HCl(aq) ---> HAuCl4(aq) + 3NO2(g) +3H2O(l)

The function of HCl in a solution of Aqua regia, that is used to dissolve gold is to dissolve other metals like quartz or iron stone that surround the gold.

Explanation: To balance the equation, we check the ratio of each element in the reacting side and the product side ( left and right hand side). Let their ratio be equal be adding moles to the compound of the element or the element it's self in either side of the equation.

HCl which is called hydrochloric acid, is an acid that does not react with gold, but it react with every other substance, like your skin, metals etc. it is used to clean a gold, by dipping the gold inside it, all the metals on the surface of the gold will dissolve.

When dissolving a gold in aqua regia solution, HCL is added to prepare this solution because it will help to dissolve all other substance on the surface of the gold.

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0.355M x 0.0282L= 0.01 moles of H2SO4. Remember sulphuric acid is diprotic so it will release 2 from each molecule. 
<span>So moles of protons = 0.01 x 2 = 0.02 moles of H+ </span>
<span>For neutralization: moles H+ = moles OH- </span>
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2 years ago
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In 1980,the US produced 18.4 billion(10to the power of 9)pounds of phosphoric acid to be used in the manufacture of fertilizer.t
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What is the amount of heat absorbed when the temperature of 75 grams of water increases from 20.°C to 35°C
SOVA2 [1]

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4702.5 J/g*k

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3 years ago
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What is the pH of a solution of 0.400 M CH₃NH₂ containing 0.250 M CH₃NH₃I? (Kb of CH₃NH₂ is 4.4 × 10⁻⁴)
Karolina [17]

Answer:

\boxed{\text{10.84}}

Explanation:

A solution of a weak base and its conjugate acid is a buffer.

The equation for the equilibrium is

\rm CH$_3$NH$_2$ + H$_2$O $\, \rightleftharpoons \,$ CH$_3$NH$_2$+ H$_{3}$O$^{+}$\\\text{For ease of typing, let's rewrite this equation as}\\\rm B + H$_2$O $\longrightarrow \,$ BH$^{+}$ + OH$^{-}$; $K_{\text{b}}$ = 4.4 \times 10^{-4}$

The Henderson-Hasselbalch equation for a basic buffer is

\text{pOH} = \text{p}K_{\text{b}} + \log\dfrac{[\text{BH}^{+}]}{\text{[B]}}

Data:

   [B] = 0.400 mol·L⁻¹

[BH⁺] = 0.250 mol·L⁻¹

    Kb = 4.4 × 10⁻⁴

Calculations:

(a) Calculate pKb

pKb = -log(4.4× 10⁻⁴)  = 3.36

(b) Calculate the pH

\text{pOH} = 3.36 + \log \dfrac{0.250}{0.400} = 3.36 + \log 0.625 = 3.36 - 0.204 = 3.16\\\\\text{pH} =14.00 -3.16 = \mathbf{10.84}\\\\\text{The pH of the solution is }\boxed{\textbf{10.84}}

4 0
2 years ago
Calculate the oxidation number of s in S2O8^2-​
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Given problem:

    S₂O₈²⁻

Find the oxidation number of S;

Oxidation number presents the extent of oxidation of each atom of elements a molecular formular or formula unit or an ionic radical.

  For radicals:

          "the algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to the charge on the ion  "

 S₂O₈²⁻;   oxidation number of O is usually -2

             2(S) + 8(-2) = -2

               2S - 16  = -2

               2S = -2 + 16

                2S  = 14

                   S  = +7

The oxidation state of S in the radical is +7

3 0
2 years ago
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