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3 years ago

Which of the following is composed of organic compounds?

2 answers:

3 0

The best answer would be HUMANS.

This is because organic compounds are essentially carbon based compounds and the human body is made up of many such compounds such as the protein that makes up skin, hair, nails, hormones and enzymes; gylcoproteins that are essential for cell membrane; fat found in skin and other tissues.

Nikolay [14]3 years ago

3 0

Answer:

Humans

Explanation:

Organic compounds are molecules that are present in living systems. These are mainly characterised by the presence of carbon (C) atoms which are linked together in the form of linear chains or arranged in ring like structures. In addition to carbon, organic compounds also contain other atoms like hydrogen (H), oxygen (O), nitrogen(N) and sulfur (S).

The human body is mainly composed of organic compounds like proteins, lipids, amino acids, sugars and carbohydrates.

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What element has an r.m.m of 72

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Explanation:

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3 years ago

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]

Answer:

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