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3 years ago

Use Hess law and the following equations to calculate H for the reactio below

Chemistry

1 answer:

aleksandr82 [10.1K]3 years ago

8 0

Answer: B2H6 (g) + 3O2 (g) → B2O3 (s) + 3H2O (g) (ΔH = -2035 kJ/mol) 3H2O (g) → 3H2O (l) (ΔH = -132 kJ/mol) 3H2O (l) → 3H2 (g) + (3/2) O2 (g) (ΔH = 858 kJ/mol)

Explanation: ??

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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 acid dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the base 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

Why is red the last color seen at sunset?

Dimas [21]

Within the visible range of light, red light waves are scattered the least by atmospheric gas molecules. So at sunrise and sunset, when the sunlight travels a long path through the atmosphere to reach our eyes, the blue light has been mostly removed, leaving mostly red and yellow light remaining.

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3 years ago

If you have 1.1 moles of magnesium nitrate then how many grams is that

marysya [2.9K]

Answer: 162.8 grams

Explanation:

Magnesium nitrate has a chemical formula of Mg(NO3)2.

Given that:

Number of moles of Mg(NO3)2 = 1.1 moles

Mass in grams of Mg(NO3)2 = ?

For Molar mass of Mg(NO3)2, use atomic mass of magnesium = 24g, nitrogen = 14g, oxygen = 16g

Mg(NO3)2 = 24g + (14g + 16gx3) x 2

= 24g + (14g + 48g) x 2

= 24g + (62g) x 2

= 24g + 124g

= 148g/mol

Now, apply the formula:

Number of moles = Mass in grams / molar mass

1.1 moles = Mass / 148g/mol

Mass = 1.1 moles x 148g/mol

Mass = 162.8 grams

Thus, there are 162.8 grams of magnesium nitrate.

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