Automobile air bags inflate following a serious

what is the pH of a solution prepared from solid, neutral 2-nitrophenol providing a fromal concentration of 0.0353M, given that

REY [17]

Answer:

pH = 4.34

Explanation:

pH= -1/2(logKa) -1/2(log C)

= -1/2( log 5.98*10^-8) -1/2(log 0.0353)

=-1/2(-7.22)-1/2(-1.45)

=3.61+0.725= 4.34

7 0

3 years ago

Which apparent motion can be explained by a geocentric model ?

baherus [9]

Answer:

The sun's path through the sky ( i hope this helps)

7 0

3 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

If 38.6 grams of iron react with an excess of bromine gas, what mass of FeBr2 can form?

Yuki888 [10]

Answer:

›› FeBr2 molecular weight. Molar mass of FeBr2 = 215.653 g/mol. This compound is also known as Iron(II) Bromide. Convert grams FeBr2 to moles or moles FeBr2 to grams. Molecular weight calculation: 55.845 + 79.904*2 ›› Percent composition by element

Explanation:

5 0

3 years ago

A sodium ion has a radius of 1.16 x 10^-10 m and a nearby fluoride ion has a radius of 1.9 x 10^-10 m. Determine the distance be

Ann [662]

The answer is: the distance between two nuclei is 2.35×10⁻¹⁰ m.

r(Na⁺) = 1.16×10⁻¹⁰ m; radius of sodium cation.

r(F⁻) = 1.9×10⁻¹⁰ m; radius of fluoride anion.

d(NaF) = r(Na⁺) + r(F⁻).

d(NaF) = 1.16×10⁻¹⁰ m + 1.9×10⁻¹⁰ m.

d(NaF) = 2.35×10⁻¹⁰ m; distance between two nuclei.

The sum of ionic radii of the cation and anion gives the distance between the ions in a crystal lattice.

4 0

4 years ago