All categories

3 years ago

Arrange these molecules in order of decreasing mass: C6H14 , NO2 , Fe2O3 , and CaCO

1 answer:

6 0

The following is the arrangement of the given molecules in decreasing order of mass: $\text{ Fe2O3} > \text{C6H14} > \text{CaCO} > \text{NO2}$

Solution:

First inorder to arrange the elements in descending order of their mass we have to calculate the molecular mass of each element. The calculation is as follows:

Mass of C6H14:

$C\rightarrow6\times12.01 = 72.06$

$H\rightarrow14\times1.008 = 14.112$

Mass of C6H14 is 86.172

Mass of NO2:

$N\rightarrow1\times14.0067 = 14.0067$

$O\rightarrow2\times15.9994 = 31.9988$

Mass of NO2 is 46.0055

Mass of Fe2O3:

$Fe\rightarrow2\times55.845 = 111.69$

$O\rightarrow3\times15.9994 = 47.9982$

Mass of Fe2O3 is 159.6882

Mass of CaCO:

$Ca\rightarrow1\times40.078 = 40.078$

$C\rightarrow1\times12.01 = 12.01$

$O\rightarrow1\times15.9994 = 15.9994$

Mass of CaCO will be 68.0874

So, the order will be $159.6882>86.172>68.0874>46.0055$ which is $\text{ Fe2O3} > \text{C6H14} > \text{CaCO} > \text{NO2}$

You might be interested in

Answer now pls and i give you mega points ;)

astraxan [27]

HCl is an acid
C5H5N is a base
Cl- is a base
HC5H5N+ is an acid

7 0

2 years ago

Finish the following word equation. calcium + nitric acid →

Dafna1 [17]

HNO3 + Ca —-> H2O + N20 + Ca(NO3)2

4 0

3 years ago

There are naturally occuring elements.

mr_godi [17]

There are 94 naturally occurring elements

6 0

3 years ago

Read 2 more answers

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

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

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 years ago

ХА

kvasek [131]

Answer:

B. Biosphere.

Explanation:

8 0

3 years ago

Arrange these molecules in order of decreasing mass: C​6​H​14​ , NO​2​ , Fe​2​O​3 ​, and CaCO​

Arrange these molecules in order of decreasing mass: C6H14 , NO2 , Fe2O3 , and CaCO