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3 years ago

How many moles of PCl5 can be produced from 24.0 g of P4

Chemistry

1 answer:

posledela3 years ago

6 0

Hey there!

Given the reaction:

P4 + 10 Cl2 ------------------ 4 PCl5

Molar mass P4 = 124 g/mol

Number of moles P4:

n = mass of solute / molar mass

n = 24.0 / 124

n = 0.1935 moles of P4

Therefore:

1 mole P4 --------------- 4 moles PCl5

0.1935 moles P4 ------- moles PCl5

moles PCl5 = 0.1935 * 4

= 0.774 moles of PCl5

Hope that helps!

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Which group or groups of atoms are the only atoms with f orbitals A. All atoms heavier than barium B. All atoms heavier than kry

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Answer:

All atoms heavier than barium

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In the periodic table, elements are divided into blocks. We have the;

s- block elements

p- block elements

d- block elements

f- block elements

However, immediately after Barium, we now encounter elements that have f-orbitals. Barium possesses a fully filled d-orbital. Hence after it, we see elements with 4f and 5f orbitals called the Lanthanides and actinides. The elements following the lanthanide and actinide series possess completely filled f-orbitals as inner orbitals.

Hence elements heavier than barium all possess f-orbitals.

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3 years ago

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Answer: The combination of element ad an ion that will react is $Ni(s)\text{ and }Pt^{2+}(aq.)$

Explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

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

For a reaction to be spontaneous, the standard electrode potential must be positive.

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

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$ ......(1)

For the given options:

Option 1: $Sn(s)\text{ and }Mn^{2+}(aq.)$

Here, tin must undergo oxidation reaction and manganese undergo reduction reaction.

Oxidation half reaction: $Sn(s)\rightarrow Sn^{2+}(aq.)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V$

Reduction half reaction: $Mn^{2+}(aq.)+2e^-\rightarrow Mn(s);E^o_{Mn^{2+}/Mn}=-1.18V$

Putting values in equation 1, we get:

$E^o_{cell}=-1.18-(-0.14)=-1.04V$

As, the standard potential is coming out to be negative, the given reaction will not take place.

Option 2: $Fe(s)\text{ and }Ca^{2+}(aq.)$

Here, iron must undergo oxidation reaction and calcium undergo reduction reaction.

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V$

Reduction half reaction: $Ca^{2+}(aq.)+2e^-\rightarrow Ca(s);E^o_{Ca^{2+}/Ca}=-2.87V$

Putting values in equation 1, we get:

$E^o_{cell}=-2.87-(-0.44)=-2.43V$

As, the standard potential is coming out to be negative, the given reaction will not take place.

Option 3: $Ni(s)\text{ and }Pt^{2+}(aq.)$

Here, nickel must undergo oxidation reaction and platinum undergo reduction reaction.

Oxidation half reaction: $Ni(s)\rightarrow Ni^{2+}(aq.)+2e^-;E^o_{Ni^{2+}/Ni}=-0.25V$

Reduction half reaction: $Pt^{2+}(aq.)+2e^-\rightarrow Pt(s);E^o_{Pt^{2+}/Pt}=1.2V$

Putting values in equation 1, we get:

$E^o_{cell}=1.2-(-0.25)=1.45V$

As, the standard potential is coming out to be positive, the given reaction will take place.

Option 4: $H_2(g)\text{ and }Na^{+}(aq.)$

Here, hydrogen must undergo oxidation reaction and sodium undergo reduction reaction.

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(aq.)+2e^-;E^o_{2H^{+}/H_2}=0V$

Reduction half reaction: $Na^{+}(aq.)+e^-\rightarrow Na(s);E^o_{Na^{+}/Na}=-0.27V$

Putting values in equation 1, we get:

$E^o_{cell}=-0.27-(-0)=-0.27V$

As, the standard potential is coming out to be negative, the given reaction will not take place.

Hence, the combination of element ad an ion that will react is $Ni(s)\text{ and }Pt^{2+}(aq.)$

8 0

3 years ago