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3 years ago

Write the ionization equation for hypochlorous acid, HClO, dissolved in water.

Chemistry

1 answer:

Pavel [41]3 years ago

5 0

Answer:

HClO (l) → H⁺ (aq) + ClO⁻ (aq)

proton and hypochlorite.

Explanation:

HClO (Hypochlorous acid)

This is a weak acid that can be dissociated as this:

HClO (l) → H⁺ (aq) + ClO⁻ (aq)

proton and hypochlorite.

It is a weak acid, so it can adopt a Ka for its equilibrium

HClO + H₂O ⇄ H₃O⁺ + ClO⁻ Ka

Molar mass = 52.46 g/m

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The pale yellow liquid bromine pentafluoride reacts violently with water to give a mixture of bromic acid and hydrofluoric acid

topjm [15]

Answer:

The concentration of bromic acid (HBrO₃) is 0.069 moles/L

the concentration of hydrofluoric acid (HF) is 0.345 moles/L

Explanation:

(a) The balanced chemical equation for the reaction is

BrF₅ + 3H₂O → HBrO₃ + 5HF

The chemical equation is balanced

(b) To determine the concentration of each acids,

First, we will determine the different volume of the acids yielded

The solution volume is 783 mL = 0.783 L

From the balanced chemical equation,

1 mole of BrF₅ produces 1 mole of HBrO₃

Likewise, 1 mole of BrF₅ will produce 5 moles of HF

From the question,

5.40×10⁻² moles of bromine pentafluoride (BrF₅) reacted;

Hence, 5.40×10⁻² moles of HBrO₃ will be produced

and 5 × 5.40×10⁻² moles of HF will be produced.

Number of moles of HBrO₃ produced = 5.40×10⁻² moles

and Number of moles of HF produced = 2.7 × 10⁻¹ moles

From the formula,

Number of moles (n) = Concentration (C) × Volume (V)

Then,

Concentration (C) = Number of moles (n) / Volume (V)

For bromic acid (HBrO₃)

n = 5.40×10⁻² moles

Volume = 0.783L

Hence,

C = 5.40×10⁻² moles / 0.783L

C = 0.069 moles/L

This is the concentration of bromic acid (HBrO₃)

For hydrofluoric acid (HF)

n = 2.7 × 10⁻¹ moles

V = 0.783L

Hence,

C = 2.7 × 10⁻¹ moles / 0.783L

C = 0.345 moles/L

This is the concentration of hydrofluoric acid (HF)

3 0

3 years ago

In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,

Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - -

At t =equilibrium (x-s) s 2s

The expression for Solubility product for CuF₂ is:

$K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2$

$K_{sp}=s\times {2s}^2$

$K_{sp}=4s^3$

Given s = 7.4×10⁻³ M

So, Ksp is:

$K_{sp}=4\times (7.4\times 10^{-3})^3$

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - 0.20

At t =equilibrium (x-s') s' 0.20+2s'

The expression for Solubility product for CuF₂ is:

$K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2$

$1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2$

Solving for s', we get

<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.</u>

3 0

4 years ago

In the united states, east coast pollutants get blown to

3241004551 [841]

Explanation:

smoking is the main reason

4 0

3 years ago

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Tatiana [17]

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5 0

3 years ago

hat is the product when magnesium reacts with nitrogen? Mg(s) + N2(g) → Mg2N3(s) Mg3N(s) Mg3N2(s) MgN3(s)

Ipatiy [6.2K]

3Mg + N₂ = Mg₃N₂

Mg₃N₂
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5 0

3 years ago

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