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3 years ago

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A mixture of two compounds, A and B, was separated by extraction. After the compounds were dried, their masses were found to be:

119 mg of compound A and 97 mg of compound B. Both compounds were recrystallized and weighed again. After recrystallization, the mass of compound A was 83 mg and the mass of compound B was 79 mg. Calculate the percent recovery from recrystallization for both compounds.

1 answer:

Arlecino [84]3 years ago

3 0

Answer:

The percent recovery from re crystallization for both compounds A and B is 69.745 and 81.44 % respectively.

Explanation:

Mass of compound A in a mixture = 119 mg

Mass of compound A after re-crystallization = 83 mg

Percent recovery from re-crystallization :

$\frac{\text{Mass after re-crystallization}}{\text{Mass before re-crystallization}}\times 100$

Percent recovery of compound A:

$\frac{83 mg}{119 mg}\times 100=69.74\%$

Mass of compound B in a mixture = 97 mg

Mass of compound B after re-crystallization = 79 mg

Percent recovery of compound B:

$\frac{79 mg}{97 mg}\times 100=81.44\%$

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Eddi Din [679]

Answer : The correct answer is Option D.

Explanation:

Ores are defined as natural occurring substance from which metals and another mineral are extracted.

Bauxite is primary ore of an aluminium metal. it doesn't have definite or specific chemical composition. This is because it contains mixture of hydrous aluminum oxide $Al_2O_3$ , aluminum hydroxides $Al(OH)_3$ , insoluble materials like hematite, quartz ,geothite ,magnetite etc.

From the given options the most appropriate statement which is correct about the bauxite is 'It does not have a definite chemical composition'.

Hence, the correct option Is (D).

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4 years ago

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topjm [15]

Answer:

8.37g

Explanation:

Step 1 :

The balanced equation for the reaction. This is given below:

N2 + 2O2 —> 2NO2

Step 2:

Data obtained from the question.

Volume (V) of N2 = 2L

Pressure (P) = 840mmHg

Temperature (T) = 24°C

Number of mole (n) of N2 =?

Step 3:

Conversion to appropriate unit.

For pressure:

760mmHg = 1atm

840mmHg = 840/760 = 1.11 atm

For Temperature:

T(K) = T(°C) + 273

T(°C) = 24°C

T(K) = 24°C + 273

T(K) = 297K

Step 4:

Determination of the number of mole N2.

The number of mole of N2 can be obtained by using the ideal gas equation as follow:

Volume (V) of N2 = 2L

Pressure (P) = 1.11 atm

Temperature (T) = 297K

Number of mole (n) of N2 =?

Gas constant (R) = 0.082atm.L/Kmol

PV = nRT

Divide both side by RT

n = PV / RT

n = 1.11 x 2 / 0.082 x 297

n = 0.091 mole

Therefore, the number of mole of N2 that reacted is 0.091 mole

Step 5:

Determination of the mass of NO2 produced from the reaction. This is illustrated below:

N2 + 2O2 —> 2NO2

From the balanced equation above,

1 mole of N2 produced 2 moles of NO2.

Therefore, 0.091 mole of N2 will produce = 0.091 x 2 = 0.182 mole of NO2.

Finally, we will convert 0.182 mole of NO2 to gram as shown below:

Number of mole NO2 = 0.182 mole

Molar mass of NO2 = 14 + (16x2) = 46g/mol

Mass = number of mole x molar mass

Mass of NO2 = 0.182 x 46

Mass of NO2 = 8.37g

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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
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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

A selectively permeable membrane separates solution a and

Yakvenalex [24]

The water molecules will flow from b to a due to osmosis.

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The net movement will stop until both sides of the solution has a same water potential.

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3 years ago

Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.89×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.25×10−4, what is the

Sergio [31]

Answer:

$1.2\times 10^{-2}$

Explanation:

The given reactions are:

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AgCl(aq)⇌Ag+(aq)+Cl−(aq) $K_4 = 1.25\times 10^{-4}$

Required reaction is:

PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(aq)

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4 0

3 years ago