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brilliants [131]

3 years ago

7

What statement is incorrect about this oxidation-reduction reaction? 2 SO2(g) + O2(g) → 2 SO3(g) What statement is incorrect abo

ut this oxidation-reduction reaction? 2 SO2(g) + O2(g) → 2 SO3(g) SO2 gains electrons. O2 is reduced. SO2 is the reducing agent. O2 is the oxidizing agent.

1 answer:

Leno4ka [110]3 years ago

6 0

Answer:

The incorrect statement is: SO₂ gains electrons

Explanation:

A chemical reaction that involves the simultaneous transfer of electrons between two chemical species, is known as the redox reaction.

Given chemical reaction: 2SO₂(g) + O₂(g) → 2SO₃(g)

In this redox reaction, S is present in +4 oxidation state in SO₂ and +6 oxidation state SO₃. Whereas, O is present in 0 oxidation state in O₂ and -2 oxidation state in SO₃.

<u>Therefore, SO₂ loses electrons and thus gets oxidized. Whereas, O₂ gains electrons and thus gets reduced. </u>

<u>In this reaction, SO₂ is the reducing agent and O₂ is the oxidizing agent.</u>

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4 years ago

Which of the changes gas to solid gas to liquid, liquid to solid liquid to gas,solid to liquid solid to gas will involve an outp

Verdich [7]

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7 0

2 years ago

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago

An aqueous magnesium chloride solution is made by dissolving 7.15 moles of MgCl2 in sufficient water so that the final volume of

irinina [24]

Answer:

2.86mol/L

Explanation:

Given parameters:

Number of moles of MgCl₂ = 7.15moles

Volume of solution = 2.50L

Unknown:

Molarity of the MgCl₂ solution = ?

Solution:

The molarity of a solution is the number of moles of solute found in a given volume.

Molarity = $\frac{number of moles }{volume}$

Insert the parameters and solve;

Molarity = $\frac{7.15}{2.5}$ = 2.86mol/L

4 0

3 years ago