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3 years ago

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What is the equation for the base ionization constant of PO4^3-

1 answer:

agasfer [191]3 years ago

3 0

Chemical reaction of PO₄³⁻ ion in water:
PO₄³⁻(aq) + H₂O(l) → HPO₄²⁻(aq) +OH⁻(aq).
Kb = [HPO₄²⁻] · [OH⁻] / [PO₄³⁻]; <span>base ionization constant.
</span>Base ionization constant <span>is the equilibrium </span>constant<span> for the </span>ionization<span> of a </span>base<span>.
</span>According to Bronsted-Lowry theory acid are donor of protons and bases are acceptors of protons (the hydrogen cation or H⁺<span>).
</span>PO₄³⁻ is Bronsted base and it can accept proton and become conjugate acid HPO₄²⁻.

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Rasheed calculates that his chemical reaction should produce 4 moles of product, but when he does the experiment, he gets only 3

Brut [27]

Answer:

The answer to your question is 75%

Explanation:

Data

Theoretical production = 4 moles

Experimental production = 3 moles

Percent yield = ?

Formula

$Percent yield = \frac{Experimental production}{Theoretical production} x 100$

Substitution

$Percent yield = \frac{3}{4} x 100$

Result

Percent yield = 75 %

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Sort these six events from the earliest to the most recent. 1. Democritus proposes the existence of atoms 3rd 2. Dalton's atomic

Nitella [24]

Explanation:

1. Democritus proposes the existence of atoms

2. Dalton's atomic theory

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4. Rutherford's gold foil experiment

5. Bohr model

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The first idea about matter containing atoms dates back to Greek philosophers. One of them was Democritus .

In 1808 Dalton put forward his atomic theory

In 1897 J.J Thomson discovered cathode rays using his gas discharge tube experiment.

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Neils Bohr in 1913 suggested his own atomic model

Erwin Schrodinger formulated the wave equation of electrons

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Dalton atomic theory brainly.com/question/1979129

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3 years ago

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Misha Larkins [42]

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An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the

san4es73 [151]

Answer:

the ionic radius of the anion $r^- = 312.52 \ pm$

Explanation:

From the diagram shown below :

The anion $Cl^-$ is located at the corners

The cation $Cs^+$ is located at the body center

The Body diagonal length = $\sqrt{3 \ a }$

∴ $2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a} \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a$

Given that :

$\frac{r^+}{r^-} =0.84$ (i.e the ratio of the ionic radius of the cation to the ionic radius of

the anion )

$0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a \\ \\ 1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a$

Also ; a = 664 pm

Then :

$r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm$

Therefore, the ionic radius of the anion $r^- = 312.52 \ pm$

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