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4 years ago

Bis(bipyridyl)cadmium(II) chloride write the chemical formula

Chemistry

2 answers:

baherus [9]4 years ago

7 0

<u>Answer:</u> The chemical formula for this complex is $[Cd(C_{10}H_8N_2)_2]Cl_2$ or $[Cd(bipy)_2]Cl_2$

<u>Explanation:</u>

2,2'-Bypridine is a bidentate chelating ligand which forms complex with many transition metals. Its chemical formula is $C_{10}H_8N_2$

Bidentate ligands are the ligands which donate two pairs of electrons to the metal atom.

This complex contains two spheres: Primary sphere and secondary sphere

Primary sphere is the sphere which satisfies the primary valency of the metal and the coordination sphere satisfies the secondary valency of the metal known as secondary sphere.

Chlorine is satisfying the primary valency of the metal and bipyridyl is satisfying the secondary valency of the metal.

The chemical formula for this complex is $[Cd(C_{10}H_8N_2)_2]Cl_2$ or $[Cd(bipy)_2]Cl_2$ and its structure is given in the image attached.

GREYUIT [131]4 years ago

3 0

Following are the chemical and structural formulae of said complex compound,

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lina2011 [118]

Answer: " 0.69 g / mL "

____________________________________________

Explanation:

______________________________________________

The "pre-lab question" given is:

_____________________________________________

"The volume of an unknown liquid is 15 ml, and the mass of the liquid and the graduated cylinder it is in is 55.2g. If the mass of the graduated cylinder is 44.8g , what is the density of the unknown liquid? " .

____________________________________________

Note: The volume of the unknown liquid is 15 mL ; regardless of whether or not the "unknown liquid" is in the graduated cylinder.

The density of the unknown liquid is measured in: "g / mL " ;

that is, "grams per mL" .

____________________________________________

Note: "D = m / V " ; that is; "Density = mass/ volume" ;

that is: ["Density = mass per 'unit volume' ]" .

____________________________________________

So; to find the density, "D" , of the "unknown liquid" ; we would have to find the "mass" of the "unknown liquid" by "subtracting" :

the "known mass of liquid when the liquid is not in the cylinder"; that is: " 44.8g" ; From:

the "known combined value of the: 'mass of the liquid PLUS the mass of the cylinder" ; that is: "55.2g" ;

→ " 55.2g - 44.8g = 10.4g " .

___________________________________________

So: " D = m / V " ; (55.2g - 44.8 g) / 15 mL " ;

= (55.2g - 44.8 g) / 15 mL " ;

= (10.4g) / 15mL ;

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= [ (10.4) / 15 } g/ mL ;

= 0.693333333333333.... g / mL ;

→ We round to "2 (two) significant figures" ;

→ since we have: "10.4 g / 15 mL " ;

and 15 mL is considered a measured/experimental value;

and: 10.4g is considered a measured/experimental value;

→ so, the least precise value; 15 mL (has only 2 (two) significant figures ;

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→ which is our answer: " 0.69 g / mL " .

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