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2 years ago

Please solve ...................................................................

Chemistry

1 answer:

scZoUnD [109]2 years ago

7 0

Answer:

Explanation:

Part (c)

Mix together two suitable solutions.

Use filtration to separate the precipitate as a residue from the solution.

Wash the precipitate with distilled water while it is in the filter funnel.

Leave the washed precipitate aside or in a warm oven to dry.

Part (d)

Pb(NO3)2 (aq) + Na2SO4 (aq) → 2 NaNO3 (aq) + PbSO4 (s)

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Let us write the appropriate equilibria and associate the correction <img src="https://tex.z-dn.net/?f=K_b" id="TexFormula1" tit

rosijanka [135]

Explanation:

The relation between $K_a\&K_b$ is given by :

$K_w=K_a\times K_b$

Where :

$K_w=1\times 10^{-14}$ = Ionic prodcut of water

The value of the first ionization constant of sodium sulfite = $K_{a1}=1.4\times 10^{-2}$

The value of $K_{b1}$ :

$1\times 10^{-14}=1.4\times 10^{-2}\times K_{b1}$

$K_{b1}=\frac{1\times 10^{-14}}{1.4\times 10^{-2}}=7.1\times 10^{-13}$

The value of the second ionization constant of sodium sulfite = $K_{a2}=6.3\times 10^{-8}$

The value of $K_{b2}$ :

$1\times 10^{-14}=6.3\times 10^{-8}\times K_{b1}$

$K_{b1}=\frac{1\times 10^{-14}}{6.3\times 10^{-8}}=1.6\times 10^{-7}$

3 0

3 years ago

You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30

dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula: $\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n$

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{1}{3})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = 0.012345$

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0

3 years ago

Given a 240.0 g sample of sulfur trioxide (MM = 80.1 g/mol),

labwork [276]

Answer:

2 mol of SO3 produces 1 mol O2

3 mol SO3 produces 3/2 mol of O2

so O2 produced = 1.5(32) =48 gm

Explanation:

8 0

2 years ago

State Hess' law of constant heat summation.

AveGali [126]

Answer:

-74.6 kj/mol

Explanation:

you can see the answer at the pic

8 0

3 years ago

A student finds a rock on the way to school. In the laboratory he determines that the volume of the rock by placing the rock in

9966 [12]

Answer:

1.76 g/mL

Explanation:

You need to find the volume. You can do this by subtracting the volume of the water and the rock by the volume of the water.

72.7 mL - 50 mL = 22.7 mL

Now that you have volume, divide the mass by the volume to find the density.

39.943 g/22.7 mL = 1.76 g/mL

7 0

3 years ago