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Sati [7]
3 years ago
15

Will a precipitate form (yes or no) when 50.0 ml of 1.2 × 10–3 m pb(no3)2 are added to 50.0 ml of 2.0 × 10–4 m na2s? if so, iden

tify the precipitate.
Chemistry
1 answer:
frez [133]3 years ago
8 0

Answer:

Yes, a precipitate of PbS forms

Explanation:

The equation of the reaction is given as:

                   Pb(NO₃)₂ + Na₂S → PbS + 2NaNO₃

The lead sulfide forms a precipitate in the aqeous solution.

Precipitation is a form of reaction in which ions combines to form a solid precipiate. Most double displacement reactions in which ionic compounds are the reactants results in formation of a precipitate as the product.

There are rules of solubility which guides a reaction that would lead to the formation of a precipitate. The mos applicable of the rules to the reaction stated above is that "carbonates, phosphates, sulfides, oxides and hydroxides are insolube". The sulfide of lead formed in the product is therefore insoluble.

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A piece of high-density Styrofoam measuring 24.0 cm by 36.0 cm by 5.0 cm floats when placed in a tub of water. When a 1.5 kg boo
zheka24 [161]

Explanation:

The given data is as follows.

         Width of Styrofoam = 24.0 cm

          Length of Styrofoam = 36.0 cm

          Height of Styrofoam = 5.0 cm

Therefore, volume of the Styrofoam will be calculated as follows.

                  Volume = length × width × height

                                =  (36.0 × 24.0 × 5.0) cm^{3}

                                 = 4320 cm^{3}

or,                             = 4.32 \times 10^{3} cm^{3}

As Styrofoam partially sinks at 3.0 cm and total height of Styrofoam is 5.0 cm. Hence, height of Styrofoam above the water is (5.0 - 3 cm) = 2 cm.

So, volume of water displaced is as follows.

          24.0 cm × 36.0 cm × 2.0 cm

         = 1.73 \times 10^{3} cm^{3}

Hence, mass of displaced water is as follows.

                 mass = density × volume

                           = 1.00 g/cm^{3} \times 1.73 \times 10^{3} cm^{3}

                           = 1.73 \times 10^{3} g

Since, book is placed on the Styrofoam. Therefore, mass of water displaced is also equal to the following.

             Mass of water displaced = mass of book + mass of Styrofoam

                  1.73 \times 10^{3} g = 1500 g + mass of Styrofoam

                   (1730 - 1500) g = mass of Styrofoam

                   mass of Styrofoam = 230 g

Therefore, calculate the density of Styrofoam as follows.

                   Density = \frac{mass}{volume}  

                                 = \frac{230}{4.32 \times 10^{3} cm^{3}}

                                 = 53.24 \times 10^{-3} g cm^{-3}

Thus, we can conclude that the density of Styrofoam is 53.24 \times 10^{-3} g cm^{-3}.

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