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3 years ago

Determine the pH of a 0.048 M hypochlorous acid (HClO) solution. Hypochlorous acid is a weak acid (Ka = 4.0 ✕ 10−8 M).

Chemistry

1 answer:

storchak [24]3 years ago

8 0

pH of 0.048 M HClO is 4.35.

Explanation:

HClO is a weak acid and it is dissociated as,

HClO ⇄ H⁺ + ClO⁻

We can write the equilibrium expression as,

Ka = $$\frac{[H^{+}] [ClO^{-}] }{[HClO]}$

Ka = 4.0 × 10⁻⁸ M

4.0 × 10⁻⁸ M = $$\frac{x \times x }{0.048}$

Now we can find x by rewriting the equation as,

x² = 4.0 × 10⁻⁸ × 0.048

= 1.92 × 10⁻⁹

Taking sqrt on both sides, we will get,

x = [H⁺] = 4.38 × 10⁻⁵

pH = -log₁₀[H⁺]

= - log₁₀[ 4.38 × 10⁻⁵]

= 4.35

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Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of

Brut [27]

Answer: The smallest temperature change is shown by water.

Explanation:

To calculate the heat absorbed or released, we use the equation:

$q=mC\times \Delta T$ ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

$\Delta T$ = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

Option a: 50.0 g Fe, $C_{Fe}=0.449J/g^oC$

We are given:

$m=50.0g\\C_{Fe}=0.449J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC$

Change in temperature = 8.99°C

Option b: 50.0 g water, $C_{water}=4.18J/g^oC$

We are given:

$m=50.0g\\C_{water}=4.18J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC$

Change in temperature = 0.96°C

Option b: 25.0 g Pb, $C_{Pb}=0.128J/g^oC$

We are given:

$m=50.0g\\C_{Pb}=0.128J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC$

Change in temperature = 62.5°C

Option d: 25.0 g Ag, $C_{Ag}=0.235J/g^oC$

We are given:

$m=25.0g\\C_{Ag}=0.235J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC$

Change in temperature = 34.04°C

Option e: 25.0 g granite, $C_{granite}=0.79J/g^oC$

We are given:

$m=25.0g\\C_{Fe}=0.79J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC$

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

5 0

3 years ago

Brainliest for an answer!

Degger [83]

The volume of H₂ : = 15.2208 L

<h3>Further explanation</h3>

Given

Reaction

2 As (s) + 6 NaOH (aq) → 2 Na₃AsO₃ (s) + 3 H₂ (g)

34.0g of As

Required

The volume of H₂ at STP

Solution

mol As (Ar = 75 g/mol) :

= mass : Ar

= 34 g : 75 g/mol

= 0.453 mol

From the equation, mol ratio As : H₂ = 2 : 3, so mol H₂ :

=3/2 x mol As

=3/2 x 0.453

= 0.6795

At STP, 1 mol = 22.4 L, so :

= 0.6795 x 22.4 L

= 15.2208 L

5 0

3 years ago

A sample of benzene was vaporized at 25◦C. When 37.5 kJ of heat was supplied, 95.0 g of the liquid benzene vaporized. What is th

grandymaker [24]

Answer:

Enthalpy of vaporization = 30.8 kj/mol

Explanation:

Given data:

Mass of benzene = 95.0 g

Heat evolved = 37.5 KJ

Enthalpy of vaporization = ?

Solution:

Molar mass of benzene = 78 g/mol

Number of moles = mass/ molar mass

Number of moles = 95 g/ 78 g/mol

Number of moles = 1.218 mol

Enthalpy of vaporization = 37.5 KJ/1.218 mol

Enthalpy of vaporization = 30.8 kj/mol

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3 years ago

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Olenka [21]

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3 years ago

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The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell

tankabanditka [31]

Explanation:

In a voltaic cell, oxidation reaction occurs at anode whereas reduction reaction occurs at the cathode.

Hence, the half-cell reaction taking place at anode and cathode will be as follows.

At anode (Oxidation) : $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$ ...... (1)

At cathode (Reduction) : $Ag^{+}(aq) + 1e^{-} \rightarrow Ag(s)$

So, in order to balance the half cell reactions, we multiply reduction reaction by 3. Hence, reduction reaction equation will be as follows.

$3Ag^{+}(aq) + 3e^{-} \rightarrow 3Ag(s)$ ........ (2)

Therefore, overall reaction will be sum of equations as (1) + (2). Thus, net reaction equation is as follows.

$Cr(s) 3Ag^{+}(aq) \rightarrow Cr^{3+}(aq) + 3Ag(s)$

6 0

3 years ago