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ZanzabumX [31]
3 years ago
8

In an ionic compound, the size of the ions affects the internuclear distance (the distance between the centers of adjacent ions)

, which affects lattice energy (a measure of the attractive force holding those ions together). Based on ion sizes, arrange these compounds by their expected lattice energy. Note that many sources define lattice energies as negative values. Please arrange by magnitude and ignore the sign.
Compunds: RbCl ,RbBr ,Rbl ,RbF
Chemistry
1 answer:
gregori [183]3 years ago
6 0

Answer:

The correct answer will be " RbF > RbCl > RbBr > Rbl".

Explanation:

The size of the given ions will be:

<u>RbCl:</u>

⇒  689kJ/mol

<u>RbBr:</u>

⇒  660kJ/mol

<u>Rbl:</u>

⇒  630kJ/mol

<u>RbF:</u>

⇒  785kJ/mol

Now according to the size, the arrangement will be:

⇒  (785kJ/mol) > (689kJ/mol) > (660kJ/mol) >(630kJ/mol)

⇒  RbF > RbCl > RbBr > Rbl

The bond among all opposite charging ions seems to be strongest whenever the ions were indeed small.

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A solution contains 3.1 mM Zn(NO3)2 and 4.2 mM Ca(NO3)2. The p-function for Zn2+ is _____, and the p-function for NO3- is _____.
seropon [69]

<u>Answer:</u> The p-function of Zn^{2+} and NO_3^{-} ions are 2.51 and 2.14 respectively.

<u>Explanation:</u>

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

Millimolar concentration of calcium nitrate = 4.2 mM

Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

  • Concentration of zinc nitrate = 0.0031 M = 0.0031 mol/L

1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, M_1=(2\times 0.0031)=0.0062M

  • Concentration of calcium nitrate = 0.0042 M = 0.0042 mol/L

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, M_2=(2\times 0.0042)=0.0084M

To calculate the concentration of nitrate ions in the solution, we use the equation:

M=\frac{M_1V_1+M_2V_2}{V_1+V_2}

Putting values in above equation, we get:

M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M

Calculating the p-function of zinc ions and nitrate ions in the solution:

  • <u>For zinc ions:</u>

\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}]

\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51

  • <u>For nitrate ions:</u>

\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}]

\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14

Hence, the p-function of Zn^{2+} and NO_3^{-} ions are 2.51 and 2.14 respectively.

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Answer: polar solvent
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