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3 years ago

Why did we use solid camphorsulfonic acid rather than aqueous h2so4?

1 answer:

3 0

Most common mineral acids solutions (HCl, H2SO4, HNO3, H3PO4, etc.) are prepared in water. Infact, system is<span> acidic only in aqueous medium.
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However, in few cases, it isn't advisable/viable to perform reaction is aqueous medium. This can be due to poor solubility of reactant in water. This situation arises especially, if reactant is highly non-polar in nature.

There also exist a possibility that, reactant is not stable in aqueous medium. In such event, use of water has to be avoided.

For such reaction, solid camphorsulfonic acid is added in reaction, when acid is required to initiate/catalyst the reaction.

The major advantage that camphorsulfonic acid offers is that, it is solid and hence it is easy to weight. Also, they don't required water addition for the reaction.

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Valence Bond theory predicts that tin will use what hybrid orbitals in Snf5 -1

LUCKY_DIMON [66]

Answer:

sp3d

Explanation:

The ground state electronic configuration of tin is written as; [Kr] 5s²4d¹⁰5p². Hybridization is a concept used to explain the combination of orbitals of appropriate energy to produce suitable orbitals that could be used for bonding.

In forming the compound Snf5^ -1, we have to hybridize the following orbitals on tin; 5p, 5d and 6s orbitals. This gives us a set of sp3d hybrid hence the answer.

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2 years ago

Why don’t scientists use the Bohr model above to explain charge transfer?

Jlenok [28]

Answer:

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2 years ago

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sasho [114]

I would say it is answer choice C, Laws are based on proven hypothesis by many different scientists.

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2 years ago

Chlorine dioxide is used as a disinfectant and bleaching agent. In water, it reacts to form chloric acid (HClO3),

Sliva [168]

All done no worries

5 0

2 years ago

Consider the following reaction:

adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

2H₂O₂ ⟶ 2H₂O + O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂

(a) Initial moles of H₂O₂

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }$

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }$

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

$\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}$

8 0

3 years ago