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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 $\text{ Fe2O3} > \text{C6H14} > \text{CaCO} > \text{NO2}$

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A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req

tigry1 [53]

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the concentration of $Cd^{2+}$ in the original solution= 0.0088 M

the concentration of $Mn^{2+}$ in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample containing Cd2+ and Mn2+ = 50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = $\frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}$

= $\frac{14.2*0.0310}{0.0600}$

= 7.3367 mL

concentration of $Cd^{2+}$ = $\frac{volume \ of \ newly \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}$

= $\frac{7.3367*0.0600}{50}$

= 0.0088 M

Thus the concentration of $Cd^{2+}$ in the original solution = 0.0088 M

Volume of excess unreacted EDTA = $\frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}$

= $\frac{16.1*0.0310}{0.0600}$

= 8.318 mL

Volume of EDTA required for sample containing $Cd^{2+}$ and $Mn^{2+}$ = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for $Mn^{2+}$ = Volume of EDTA required for

sample containing $Cd^{2+}$ and

$Mn^{2+}$ -- Volume of newly freed EDTA

Volume of EDTA required for $Mn^{2+}$ = 55.682 - 7.3367

= 48.3453 mL

Concentration of $Mn^{2+}$ = $\frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}$

Concentration of $Mn^{2+}$ = $\frac{48.3453*0.0600}{50}$

Concentration of $Mn^{2+}$ in the original solution= 0.058 M

Thus the concentration of $Mn^{2+}$ = 0.058 M

6 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