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Zolol [24]
2 years ago
12

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 doe

s 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 HNO, and HC in a 1:3 ratio
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 from the acid solution 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. Add the stoichiometric coefficients to the equation to balance it.
equation:Au(s)+ HNO, (ag)+HClagHAuCI(a)+No,(g)+H,00)
What is the function of HCI?
A) It acts as a reducing agent.
B) It supplies chloride ions to form a complex ion with the oxidized gold.
C) It acts as an oxidizing agent.
Chemistry
1 answer:
Paul [167]2 years ago
4 0

The equation given is incorrect, the correct equation is:

Au(s) + HNO₃(aq) + HCl(aq) → HAuCl₄(aq) + NO₂(g) + H₂O(l)

Answer:

B) It supplies chloride ions to form a complex ion with the oxidized gold.

Explanation:

The equation given is:

Au(s) + HNO₃(aq) + HCl(aq) → HAuCl₄(aq) + NO₂(g) + H₂O(l)

To balance the equation, we need to find the oxidation number (Nox) of the elements, and identify which substance is being oxided and which on is being reduced. Simple elements has Nox = 0, H has Nox = +1, O has Nox = -2. So:

Au(s): Nox Au = 0

HNO₃(aq): Nox H = +1; Nox O = -2, Nox N: +1 +x +3*(-2) = 0 -> x = +5

HCl(aq): Nox H = +1, Nox Cl = -1

HAuCl₄(aq): Nox H = +1, Nox Cl = -1, Nox Au: +1 +x +4*(-1) = 0 -> x = +3

NO₂(g): Nox O = -2; Nox N: x + 2*(-2) = 0 -> x = +4

H₂O(l): Nox H = +1; Nox O = -2

So, Au is being oxidezed from 0 to +3, and N is being reduced from +5 to +4:

ΔNox(Au) = 3

ΔNox(N) = 1

Both of them are single in the compound they are, so it's not necessary to multiply by the coefficient (because is 1). So, the ΔNox must be changed between the compounds:

Au(s) + 3HNO₃(aq) + HCl(aq) → HAuCl₄(aq) + NO₂(g) + H₂O(l)

Now, the balancing must be done by trial, knowing that the elements must be at the same number on both sides of the equation. So, we multiply HCl by 4, and NO₂ and H₂O by 3:

Au(s) + 3HNO₃(aq) + 4HCl(aq) → HAuCl₄(aq) + 3NO₂(g) + 3H₂O(l).

So, we can see that Au is being oxidized, so it's the reducing agent, and HNO₃ ins being reduced, so it's the oxidizing agent. The HCl supplies Cl ions to the complex formed with gold.

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A chemist mixes 75.0 g of an unknown substance at 96.5C w/ 1,150 g of water at 25.0C . if the final temperature of the system is
vichka [17]
Remember: heat lost = heat gained 

When calculating heat loss or gain, remember 

mass*(spec heat cap)*(change in T) 

The unknown loses heat- we don't know the spec heat cap, so we'll call it x.

The water gains. I've omitted the units, but always use when solving problems on your own. 

75*x*(96.5-37.1) = 1150*4.184*(37.1-25) 
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Now it's all set up- use algebra to get x, the spec heat cap of the unk in J/g*degC

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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3 0
2 years ago
A solution has a hydroxide-ion concentration of 0.0040 M. What is the pOH of the solution? A solution has a pH value of 3.66. Wh
hodyreva [135]

Answer:

1. pOH = 2.4

2. pOH = 10.34

3. pH = 2.14

Explanation:

1. Determination of the pOH.

Concentration of Hydroxide ion, [OH-] = 0.004 M

pOH =?

pOH = - log [OH-]

pOH = - log (0.004)

pOH = 2.4

Therefore, the pOH of the solution is 2.4

2. Determination of the pOH.

pH = 3.66

pOH =?

pH and pOH are related by the following equation:

pH + pOH = 14

With the above formula, we can obtain the pOH of the solution as follow:

pH = 3.66

pOH =?

pH + pOH = 14

3.66 + pOH = 14

Collect like terms

pOH = 14 - 3.66

pOH = 10.34

Therefore, the pOH of the solution is 10.34

3. Determination of the pH.

Molarity of HCl = 0.0072 M

Concentration of Hydrogen ion, [H+] =?

Thus, we can obtain the concentration of Hydrogen ion, [H+] as follow:

HCl(aq) —> H+(aq) + Cl-(aq)

From the balanced equation above,

1 mole of HCl produced 1 mole of H+.

Therefore, 0.0072 M HCl will also produce 0.0072 M H+.

Therefore, the concentration of Hydrogen ion, [H+] in the solution is 0.0072 M.

Finally, we shall determine the pH of the solution as follow:

Concentration of Hydrogen ion, [H+] = 0.0072 M.

pH =?

pH = - log [H+]

pH = - log (0.0072)

pH = 2.14

Therefore, the pH of the solution is 2.14

3 0
2 years ago
A sample of 0.6760 g of an unknown compound containing barium ions (ba2+) is dissolved in water and treated with an excess of na
notka56 [123]

Answer: 35.72 % of Barium ions will be present in the original unknown compound.

Explanation: The reaction of Barium ions and sodium sulfate is:

Na_2SO_4(aq.)+Ba^{2+}(aq.)\rightarrow BaSO_4(s)+2Na^+(aq.)

Here, Sodium sulfate is present in excess, Barium ions are the limiting reagent because it limits the formation of product.

Now, 1 mole of barium sulfate is produced by 1 mole of Barium ions.

Molar mass of Barium sulfate = 233.38 g/mol

Molar mass of Barium ions = 137.327 g/mol

233.38 g/mol of barium sulfate will be produced by 137.323 g/mol of Barium ions, so

0.4105 grams of barium sulfate will be produced by = \frac{137.327g/mol}{233.38g/mol}\times 0.4105g of Barium ions

Mass of barium ions = 0.2415 grams

To calculate percentage by mass, we use the formula:

\% mass=\frac{\text{Mass of solute (in grams)}}{\text{Total mass of the solution(in grams)}}\times 100

Mass of the solution = 0.6760 grams

Putting the value in above equation, we get

\% \text{ mass of }Ba^{2+}\text{ ions}=\frac{0.2415g}{0.6760g}\times 100

% mass of Barium ions = 35.72%.

8 0
2 years ago
Be sure to answer all parts. calculate δg o for the reaction between i2(s) and br−(aq). e o cell = −0.54 j/c enter your answer i
denpristay [2]

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

The attached file have the solved problem.

3 0
2 years ago
What is the temperature (in K) of 16.45 moles of methane gas in a 4.95 L container at 4.68 atm?
AlexFokin [52]
Using the ideal gas law: PV=nRT
      P is pressure; V is volume; n is the amount in moles; R=0.082; T is temperature in K.

(4.68)*(4.95)=(16.45)*(0.0821)*T
Solve for T. 
T=17.15
4 0
2 years ago
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