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3 years ago

A sample of gold (Au) has a mass of 45.39 g.

1 answer:

8 0

1. What is the molar mass of gold?
Molar mass is a unit that expresses the mass of a molecule per one mol. The molar mass can be obtained by adding the neutron with the proton of the atoms. Gold has atomic number 79 so the proton is 79. The number of the neutron is 118. Then the molar mass would be: 79 + 118 = 197 g/mol

2. Calculate the number of moles of gold (Au) in the sample. Show your work.
In this question, you are given the mass of the gold and asked for how many moles the sample has. To find the number of moles you just need to divide the weight by the molar mass.
For 45.39 grams of gold, the number of moles would be:
45.39 / (197g/mol)= 0.23 moles

3. Calculate the number of atoms of gold (Au) in the sample. Show your work.Moles is unit of a number of molecules but 1 mol doesn't represent 1 molecule. The number of atoms can be obtained by multiplying the number of moles with Avogadro number. The calculation would be:
0.23 moles * (6.023 * 10^23 molecules/mol)= 1.387 * 10^23 molecules

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Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

draw the lewis structure, including unshared pairs, of the following molecule. carbon has four bonds in the compound. propane (c

SIZIF [17.4K]

There are three carbon and 8 hydrogen present in propane molecule. The Lewis structure of propane is shown as:

Three molecules of such carbon atoms bound to eight molecules with hydrogen atoms make up the organic complex propane molecule.

It is known that carbon has 4 valence electrons and hydrogen has one valence electron. Carbon needs 4 extra electrons to complete its octet hence, it will share its electrons with with 4 hydrogen atom and complete its octet.

Carbon will be formed 4 bond . Three bond with hydrogen and one bond with carbon atom.

To know more about Lewis structure.

brainly.com/question/15837141

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2 years ago

What became of alchemy after the 1700s

Dvinal [7]

Chemistry developed from alchemy after the 1700s. It was the Alchemist observations and accidental discoveries that brought around modern chemistry.

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3 years ago

One should know properties of constituents of a mixture to separate the mixture .Why?

klio [65]

Answer:

Because: The mixtures contain unwanted substances which may be harmful and may degrade the properties of mixtures. So we, need to separated them and extract useful substances.

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3 years ago

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Among the elements of the main group, the general trend in the first ionization energy moving across a period is not followed be

Radda [10]

The Ionization energy decreases because the full s orbital shields the electron entering the p orbital. This is known as "shielding", When each new electron experiences attraction from the nucleus and repulsion forces from the S orbitals, the net force on outer shell electrons is significantly smaller. Therefore, ionization energy decreases during these groups.

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3 years ago

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