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stich3 [128]
1 year ago
8

Arrange the following substances in order of decreasing magnitude of lattice energy. Rank the compounds in order of decreasing m

agnitude of lattice energy.a. MgOb. CaOc. NaFd. KCl
Chemistry
1 answer:
True [87]1 year ago
3 0

The lattice energy of the compounds is distributed in the following decreasing order of magnitude: MgO > CaO > NaF > KCl.

<h3>KCl or NaF, which has a higher lattice energy?</h3>

The lattice energy increases with increasing charge and decreasing ion size.(Refer to Coulomb's Law.)MgF2 > MgO.Following that, we can examine NaF and KCl (both of which have 1+ and 1-charges), as well as atomic radii.NaF will have a larger LE than KCl since Na is smaller then K and F was smaller than Cl.

<h3>MgO or CaO, which has a larger lattice energy?</h3>

MGO is more difficult than CaO, hence.This is because "Mg" (two-plus) ions are smaller than "Ca" (two-plus) ions in size.MgO has higher lattice energy as a result.

To know more about compounds visit:

brainly.com/question/14117795

#SPJ4

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The normal freezing point of a certain liquid
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Explanation :  Given,

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Formula used :  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}\times 1000}{\text{Molar mass of urea}\times \text{Mass of X liquid}}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution = -0.5^oC

\Delta T^o = freezing point of liquid X= 0.4^oC

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K_f = molal freezing point depression constant of X = ?

m = molality

Now put all the given values in this formula, we get

[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}

k_f=4.12^oC/m

Therefore, the molal freezing point depression constant of X is 4.12^oC/m

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