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.
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Answer:
sand and water
Explanation:
when u mix these both it will become like a paste
Answer: 8, which is basic
Explanation: The hydrogen ion concentration is 1.0 x 10^-8 mole per liter. Using [H+] = 1.0 x 10 -(pH) the pH is 8. A pH above 7 is basic.
Answer:
118750 ml
Explanation:
The chemical equation for complete combustion of hexane is given as;
2C6H14 + 19O2 → 12CO2 + 14H2O
From the equation of the reaction;
2 mol of C6H14 reacts with 19 mol of O2
2 ml of C6H14 reacts with 19 ml of O2
2500 mL of C6H14 would react with x ml of O2
2 = 19
2500 = x
x = 2500 * 19 / 2 = 23750 ml
Since oxygen is 20% of air;
23750 = 20 / 100 * (Volume of air)
Volume of air = 23750 * 100 / 20 = 118750 ml
The balanced chemical reaction would be as follows:
<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10
We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:
8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2
Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:
0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>
