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2 years ago

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Using the reaction C12H22O11 + 12 O2 ->12CO2 + 11H2O (e). How many grams of carbon dioxide can be produced from 456.7g sucros

e

1 answer:

klio [65]2 years ago

7 0

Answer:

704.6 g CO2

Explanation:

MM sucrose = 342.3 g/mol

MM CO2 = 44.01 g/mol

g CO2 = 456.7 g sucrose x (1 mol sucrose/MM sucrose) x (12 moles CO2/1 mol sucrose) x (MM CO2/1mol CO2) = 704.6 g CO2

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Consider the equation: 2NO2(g) N2O4(g). Using ONLY the information given by the equation which of the following changes would in

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3 years ago

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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 $pK_b$ :

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

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

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

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

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

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3 years ago

Which is the strongest acid listed in the table?

andrew11 [14]

Answer:

Hydrofluoric acid.

Explanation:

To know which of the acid is the strongest, let us determine the pka of each acid. This is illustrated below:

1. Acetic acid

Ka = 1.8x10^-5

pKa =..?

pKa = –logKa

pKa = –Log 1.8x10^-5

pKa = 4.74

2. Benzoic acid

Ka = 6.5x10^-5

pKa =..?

pKa = –logKa

pKa = –Log 6.5x10^-5

pKa = 4.18

3. Hydrofluoric acid.

Ka = 6.8x10^-4

pKa =..?

pKa = –logKa

pKa = –Log 6.8x10^-4

pKa = 3.17

4. Hypochlorous acid

Ka = 3.0x10^-8

pKa =..?

pKa = –logKa

pKa = –Log 3.0x10^-8

pKa = 7.52

Note: the smaller the pKa value, the stronger the acid.

The pka of the various acids as calculated above is given below:

Acid >>>>>>>>>>>>>>>>>> pKa

1. Acetic acid >>>>>>>>>> 4.74

2. Benzoic acid >>>>>>>> 4.18

3. Hydrofluoric acid >>>> 3.17

4. Hypochlorous acid >> 7.52

From the above illustration, we can see that hydrofluoric acid has the lowest pKa value. Therefore, hydrofluoric acid is the strongest among them.

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3 years ago

Which statement describes characteristics of earthquakes occurring along divergent plate boundaries?

IgorC [24]

B
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