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Luda [366]
4 years ago
10

Consider again the solution made by diluting 10.0 mL of the 0.50 M NaCl by adding distilled water until 100 mL of solution are p

roduced.
Has the concentration of the diluted salt solution increased, decreased, or remained constant?
Chemistry
2 answers:
alexandr1967 [171]4 years ago
8 0

Answer:

Decreased

Explanation:

Using the dilution equation:

C₁V₁ = C₂V₂

where C₁ and V₁ are concentration and volume before dilution and C₂ and V₂ concentration and volume after dilution

Plugging the numbers we have,

0.5 M x 10 ml = C₂ x 100 ml

C₂ = (0.5 M x 10 ml) / 100 ml = 0.05 M

Therefore, the concentration of the diluted salt solution has decreased.

Darina [25.2K]4 years ago
5 0
Every time you add water to a solution the concentration decreases<span />
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3 years ago
Use the born-haber cycle to calculate the lattice energy of kcl. (δhsub for potassium is 89.0 kj/mol, ie1 for potassium is 419 k
eduard

Given data:

Sublimation of K

K(s) ↔ K(g)                            ΔH(sub) = 89.0 kj/mol

Ionization energy for K

K(s) → K⁺ + e⁻                         IE(K) = 419 Kj/mol

Electron affinity for Cl

Cl(g) + e⁻ → Cl⁻                      EA(Cl) = -349 kj/mol

Bond energy for Cl₂

1/2Cl₂ (g) → Cl                        Bond energy = 243/2 = 121.5 kj/mol

Formation of KCl

K(s) + 1/2Cl₂(g) → KCl(s)        ΔHf = -436.5 kJ/mol

<u>To determine:</u>

Lattice energy of KCl

K⁺(g) + Cl⁻(g) → KCl (s)                   U(KCl) = ?

<u>Explanation:</u>

The enthalpy of formation of KCl can be expressed in terms of the sum of all the above processes, i.e.

ΔHf(KCl) = U(KCl) + ΔH(sub) + IE(K) + 1/2 BE(Cl₂) + EA(Cl)

therefore:

U(KCl) = ΔHf(KCl) - [ΔH(sub) + IE(K) + 1/2 BE(Cl₂) + EA(Cl)]

         = -436.5 - [89 + 419 + 243/2 -349] = -717 kJ/mol

Ans: the lattice energy of KCl = -717 kj/mol



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