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Fudgin [204]
3 years ago
9

Write the formula of the coordination compound pentaamminecarbonatocobalt(III) iodide. Enclose the coordination complex in squar

e brackets, even if there are no counter ions. Do not enclose a ligand in parentheses if it appears only once. Enter water as H2O. Use any abbreviation given in blue below, and enclose it in parentheses if there is more than one.
Chemistry
1 answer:
7nadin3 [17]3 years ago
6 0

Answer:

[Co(NH3)5CO3]I3

Explanation:

The naming of coordination compounds follows certain rules specified by IUPAC. Usually, the name of the complex makes it quite easy to deduce its structure.

"Pentaamine" means that there are five NH3 ligands as shown in the structure. The ligand carbonato is CO3^2-.  It has no prefix attached to it in the IUPAC name of the complex hence there is only one carbonato ligand present(recall that the complex has a coordination number of six). I did not enclose it within parenthesis as required in the question.

Lastly the III that appeared after the metal name "cobalt" shows its oxidation state. The iodide counter ions must then be 3 in number in order to satisfy this primary valency of the metal hence the inclusion of I3 in the structure of the complex.

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You are on an alien planet where the names for substances and the units of measures are very unfamiliar.
neonofarm [45]

Given that in the alien planet the names of substances and the units in which they are measured in not familiar.

8 quibs of a substance called skvarnik is present.

The conversion factor between sleps and quibs is: 9 sleps = 13 quibs

Converting 8 quibs of skvarnick to sleps:

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3 years ago
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Alekssandra [29.7K]

<u>Answer: </u>The molecular weight of the dibasic acid is 89.6 g/mol

<u>Explanation:</u>

Normality is defined as the amount of solute expressed in the number of gram equivalents present per liter of solution. The units of normality are eq/L. The formula used to calculate normality:

\text{Normality}=\frac{\text{Given mass of solute}\times 1000}{\text{Equivalent mass of solute}\times \text{Volume of solution (mL)}}      ....(1)

We are given:

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Given mass of solute = 0.56 g

Volume of solution = 250 mL

Putting values in equation 1, we get:

0.05=\frac{0.56\times 1000}{\text{Equivalent mass of solute}\times 250}\\\\\text{Equivalent mass of solute}=\frac{0.56\times 1000}{0.05\times 250}=44.8g/eq

Equivalent weight of an acid is calculated by using the equation:

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Putting values in equation 2, we get:

44.8g/eq=\frac{\text{Molar mass}}{2eq/mol}\\\\\text{Molar mass}=(44.8g/eq\times 2eq/mol)=89.6g/mol

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