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Otrada [13]
3 years ago
6

Consider two solutes, Solute A and Solute B, that have the same lattice energy. Both solutes dissolve exothermically in water, b

ut Solute A creates weaker ion-dipole intermolecular forces with water. Which solute has a more exothermic dissolution overall?
Chemistry
1 answer:
frosja888 [35]3 years ago
8 0

Consider two solutes, Solute A and Solute B, that have the same lattice energy. Both solutes dissolve exothermically in water, but Solute A creates weaker ion-dipole intermolecular forces with water. Which solute has a more exothermic dissolution overall

Answer:

Solute B has a more exothermic dissolution

Explanation:

Since Solute A and Solute B have same lattice energy but solution A has a weaker ion-dipole intermolecular forces with water which is the solvent

Solution B will dissolve faster due to higher ion-dipole intermolecular force compare to solution B, in the process of dissolve release energy to it solvent.

Therefore Solute B has a more exothermic dissolution

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A 65.0-gram sample of some unknown metal at 100.0° C is added to 100.8 grams of water at 22.0° C. The temperature of the water r
vampirchik [111]
We do a heat balance to solve this:
(m cp ΔT)water = -(m cp ΔT)metal
100.8 (4.18) (27 - 22) = -65 (cp)(27-100)
cp = 100.8 (4.18) (27 - 22) / (-65 (27-100))
cp = 0.44 J/ (°C × g)

The specific heat of the metal is 0.44 J/ (°C × g)
5 0
2 years ago
A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)↽−−⇀H+(aq)+A−(aq) The equilibrium concentratio
Pachacha [2.7K]

Answer:

pK_{a} of HA is 6.80

Explanation:

pK_{a}=-logK_{a}

Acid dissociation constant (K_{a}) of HA is represented as-

                K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}

Where species inside third bracket represents equilibrium concentrations

Now, plug in all the given equilibrium concentration into above equation-

K_{a}=\frac{(2.00\times 10^{-4})\times (2.00\times 10^{-4})}{0.250}

So, K_{a}=1.6\times 10^{-7}

Hence pK_{a}=-log(1.6\times 10^{-7})=6.80

6 0
2 years ago
How many chlorine ions are required to bond with one aluminum ion and why??
antiseptic1488 [7]
3 Chlorine ions are required to bond with one aluminum ion.

In ionic bonds, metals atoms loses all its outermost shell electrons to form a cation. While, non metal atoms gains however many electrons in order to make its outermost electron shell be 8 (or 2 if there's only one shell).

Therefore, form the periodic table, we can see that aluminum has a atomic number of 13, which makes its electron arrangement be 2,8,3. So, in order to form a aluminum ion, an Al atom must lose 3 electrons. On the other hand, Chlorine has a atomic number of 17, which means it has the electron configuration of 2,8,7. It has to gain only 1 electron to have 8 outermost shell electron.

Thereofre, 3 Chlorine atom are required to gain all 3 electrons given out by just 1 aluminum ion.
8 0
3 years ago
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grin007 [14]

Ionic compounds are formed by the complete transfer of electrons from more electronegative elements to less electronegative elements.

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9. The <u>compound SrF₂</u>

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An ionic compound is a chemical compound composed of ions held together via electrostatic forces termed ionic bonding. The compound is neutral standard but includes definitely charged ions known as cations and negatively charged ions known as anions.

An ionic compound is a chemical compound composed of ions held collectively by means of electrostatic forces termed ionic bonding. The compound is neutral normal however consists of positively charged ions known as cations and negatively charged ions referred to as anions. Ionic compounds incorporate ions and are held collectively via the attractive forces of most of the oppositely charged ions.

Learn more about ionic compounds here:-

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