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taurus [48]
3 years ago
14

Why is gold foil used on sculptures

Chemistry
2 answers:
Troyanec [42]3 years ago
7 0
Its to give it appeal painting the same would discolour quicker.
Aleonysh [2.5K]3 years ago
5 0

The gold leaf (not foil) is to give a sculpture or a painting appeal. Without it, it would discolour quicker.The gold leaf can be easily removed if needed to be.

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Alchemists in the Middle Ages dreamed of converting base metals, such as lead, into precious metals—gold and silver. Why could t
lesantik [10]

Explanation:

Each element in the periodic table has different but fixed number of the protons in nucleus of it's atom, which is known as the atomic number.

Transmutation of one chemical element into the another involves the changing of the atomic number. Such nuclear reaction requires millions of the times more energy as compared to normal chemical reactions. Thus, the dream of  the alchemist of transmuting the lead into the gold was never achievable chemically .

Conversion of lead to gold in today's world:

This conversion is indeed possible. The requirements are a particle accelerator, tremendous supply of the energy. Nuclear scientists at the Lawrence Berkeley National Laboratory located in California, more than 30 years ago, succeeded in producing very minute amounts of the gold from the bismuth. Bismuth is a metallic element which is adjacent to the lead on periodic table. Same process would work for the lead but isolating gold at end of reaction would prove much more difficult because lead is available in many isotopes. The homogeneous nature of the element means that it is easier to separate the gold from the bismuth as compared to separate the gold from the lead which has four  isotopic identities which all are stable.

4 0
3 years ago
Where does the equilibrium point occur in a reaction system?
Ymorist [56]
Equilibrium occurs when the rate of the forward reaction is the same as the rate of the reverse reaction. This doesn't necessarily mean the concentrations or pressure are the same on both sides of the equation, only the rates are the same
8 0
3 years ago
What is the chemical name for the compound with the formula Na2S?
Marina86 [1]
I think it's Sodium Sulfide, because Na2S contains one Sodium and two Sulfur. Plus Sulfide<span> is used to describe any of three types of chemical compounds that contain sulfur.</span> 
3 0
4 years ago
Read 2 more answers
Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --
frosja888 [35]

Answer :

(a) The anode and cathode will be E^o_{(Cu^{2+}/Cu)}

and E^o_{(Fe^{3+}/Fe^{2+})} respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  Fe^{3+}+1e^-\rightarrow Fe^{2+}

Thus, the anode and cathode will be E^o_{(Cu^{2+}/Cu)}

and E^o_{(Fe^{3+}/Fe^{2+})} respectively.

(b) Now we have to calculate the potential of a cell.

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  Fe^{3+}+1e^-\rightarrow Fe^{2+}

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}

The overall cell reaction will be,

2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}

E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V

E^o_{[Cu^{2+}/Cu]}=0.34V

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}

E^o=1.54V-(0.34V)=1.20V

Now we have to calculate the cell potential.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = emf of the cell = ?

Now put all the given values in the above equation, we get:

E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}

E_{cell}=1.022V

Therefore, the emf of cell potential is 1.022 V

5 0
3 years ago
A chemical change occurs when select one:
Ne4ueva [31]
I think it is d. Because the process is not reversible and it a chemical process
7 0
3 years ago
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