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astra-53 [7]
3 years ago
5

Like all equilibrium constants, the value of kw depends on temperature. at body temperature (37∘c), kw=2.4⋅10−14. part a what is

the [h3o+] in pure water at body temperature?
Chemistry
1 answer:
LenaWriter [7]3 years ago
6 0
First, we have to know that we can put [H+] instead of [H3O+]

so, according to the reaction equation:

by using ICE table:

             H2O ↔  H+  +  OH-

initial                    0           0

change                +X         +X

Equ                       X             X

when Kw = [H+] [OH-]

and when we have Kw = 2.4 x 10^-14 

and when [H+] = [OH-] = X

∴ 2.4 x 10^-14 = X^2

∴ X = √(2.4 x 10^-14)

      = 1.55 x 10^-7

∴[H+] = 1.55 x 10^-7


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3 0
3 years ago
Given the following thermodynamic data, calculate the lattice energy of LiCl:
tiny-mole [99]

Answer:

\boxed{\text{-862 kJ/mol}}

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One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1)  Li(s) + ½Cl₂ (g) ⟶ LiCl(s);      ΔHf°     = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g);                          ΔHsub =    161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g)                     BE        =   243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻                   IE₁         =   520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g)                EA₁       =  -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

                                                <u>E/kJ </u> 

(5) Li⁺(g) +e⁻ ⟶ Li(g)                520

(6) Li(g) ⟶ Li(s)                         -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s)     -409

(8) Cl(g) ⟶ ½Cl₂(g)                   -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻               <u>+349</u>

      Li⁺(g) +  Cl⁻(g) ⟶ LiCl(s)     -862

The lattice energy of LiCl is \boxed{\textbf{-862 kJ/mol}}.

3 0
3 years ago
Can anybody help me with these 2 ohms law
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yywhhwhwhwysuiwjwcwcwfwhwuwiwioowow

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