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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.

You might be interested in
How many hydrogen bonds can CH2O make to water
VladimirAG [237]
Hydrogen bonds are not like covalent bonds. They are nowhere near as strong and you can't think of them in terms of a definite number like a valence. Polar molecules interact with each other and hydrogen bonds are an example of this where the interaction is especially strong. In your example you could represent it like this: 

<span>H2C=O---------H-OH </span>

<span>But you should remember that the H2O molecule will be exchanging constantly with others in the solvation shell of the formaldehyde molecule and these in turn will be exchanging with other H2O molecules in the bulk solution. </span>

<span>Formaldehyde in aqueous solution is in equilibrium with its hydrate. </span>

<span>H2C=O + H2O <-----------------> H2C(OH)2</span>
5 0
2 years ago
Calculate the mass in grams of each of the following amounts: 1.002 mol of chromium 4.08 x 10-8 mol of neon
Pepsi [2]

Answer:

Mass_{chromium}=52.1\ g

Mass_{neon}=8.23\times 10^{-7}\ g

Explanation:

<u>Calculation of the mass of chromium as:- </u>

Moles = 1.002 moles

Molar mass of chromium = 51.9961 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

1.002\ mol= \frac{Mass}{51.9961\ g/mol}

Mass_{chromium}=1.002\times 51.9961\ g = 52.1\ g

<u>Calculation of the mass of neon as:- </u>

Moles = 4.08\times 10^{-8} moles

Molar mass of neon = 20.1797 g/mol

Thus,

1.002\ mol= \frac{Mass}{20.1797\ g/mol}

Mass_{neon}=4.08\times 10^{-8}\times 20.1797\ g = 8.23\times 10^{-7}\ g

6 0
2 years ago
How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C
NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

3 0
2 years ago
When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C
UNO [17]

Answer:

The new molar concentration of CO at equilibrium will be  :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M

Equilibrium constant of the reaction :

K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}

K = 4

CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)

Concentration at eq'm:

0.24 M          0.24 M                 0.48 M            0.48 M

After addition of 0.34 moles per liter of CO_2 and H_2 are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}

4=\frac{(0.48+x)^2}{(0.24+0.34)^2}

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

3 0
2 years ago
For the reaction below, Kp 5 1.16 at 800.8C. CaCO3(s) 34 CaO(s) 1 CO2(g) If a 20.0-g sample of CaCO3 is put into a 10.0-L contai
Elena L [17]

Answer:

The mass percentage of calcium carbonated reacted is 2.5%.

Explanation:

The reaction is:

CaCO_{3}(s)--->CaO(s)+CO_{2}(g)

Thus the Kp of the equilibrium will be:

Kp = partial pressure of carbon dioxide [as the other are solid]

Moles of calcium carbonate initially present = \frac{mass}{molarmass}=\frac{20}{100}=0.2

Let us apply ICE table to the equilibrium given:

                        CaCO_{3}(s)--->CaO(s)+CO_{2}(g)

Initial                       0.2                       0          0

Change                 -x                            +x        +x

Equilibrium           0.2-x                         x          x

Kp = partial pressure of carbon dioxide

Kp = Kc(RT)ⁿ

where n = difference in the number of moles of gaseous products and reactants

for given reaction n = 1

R = gas constant = 8.314 J /mol K

T = temperature = 800 ⁰C = 1073 K

Putting values

Kc =\frac{Kp}{RT}=\frac{1.16}{8.314X1073}=1.3X10^{-4}

Kc = \frac{[CO_{2}][CaO]}{[CaCO_{3}]}= \frac{x^{2} }{(0.2-x)}=1.3X10^{-4}

1.3X10^{-4}(0.2-x)=x^{2}

x^{2} = 0.26X10^{-4}-1.3X10^{-4}x

On calculating

x =  0.005

where x = the moles of calcium carbonate dissociated or reacted.

Percentage of the moles or mass reacted = \frac{molesreacted X100}{initialmoles}=\frac{0.005X100}{0.2}=2.5%

7 0
2 years ago
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