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bagirrra123 [75]
2 years ago
8

When a drawing of a circle made in the center of a piece of paper with a black marker gets wet, the marker bleeds, and as the wa

ter moves through the paper the ink separates into several colors.
Chemistry
1 answer:
Ahat [919]2 years ago
8 0

Answer:

Chromatography

Explanation:

Chromatography is a technique for separating mixtures that involves the use of a moving liquid and filter paper. The solvent travels through the spots on the paper, taking the chemicals away in the opposite direction. If the solvent combination has been properly chosen, each of them will move at a different velocity than the others.

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An alkaline battery produces electrical energy according to the following equation.
Gnesinka [82]

Answer:

Part a: limiting reactant MnO₂

Part b: 12.43 g of Zn(OH)₂

Explanation:

Part a

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Limiting reactant = ?

Given Reaction

        Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

To determine the limiting reactant first we will look at the reaction

         Zn(s)  +  2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

          1 mol     2 mol

Convert moles to mass

molar mass of Zn = 65.4 g/mol

Mass of MnO₂ = 55 + 2 (16) = 55 + 32 = 87 g/mol

So,

       Zn(s)          +      2 MnO₂(s)    +    H₂O(l) ----------> Zn(OH)₂(s) + Mn₂O₃(s)

1 mol (65.4 g/mol)    2 mol (87 g/mol)

       65 g                        174 g

So its clear from the reaction that 65 g Zn react with 174 g of MnO₂.

now if we look at the given amounts the amount MnO₂ is less then the amount of Zn but in actual calculation amount of MnO₂ is more then amount of zinc.

So, for MnO₂ if we calculate the needed amount of zinc

So apply unity formula

           65 g Zn react ≅ 174 g of MnO₂

            X g of Zn ≅ 21.5 g of MnO₂

Do cross multiplication

           X g Zn react = 65 g x 21.5 g / 174 g

           X g of Zn ≅ 8.032 g

So, 8.032 g of zinc will react out of 37.0 grams. the remaining will be in excess.

So MnO₂ will be consumed completely an it will be limiting reactant.

____________

part b

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Mass of Zn(OH)₂  produced = ?

Given Reaction

        Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

Solution:

As from the part A we come to know that MnO₂ is limiting reactant, so the amount of Zn(OH)₂ will depend on the amount of MnO₂.

So first we convert mass of MnO₂ to moles

Formula Used

        no. of moles = mass in grams / molar mass

Mass of MnO₂ = 55 + 2 (16)

Mass of MnO₂ = 55 + 32 = 87 g/mol

Put values in the above equation

        no. of moles = 21.5 g / 87 g/mol

        no. of moles = 0.25 moles

Now,

Look at the reaction

         Zn(s)  +  2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

                        2 mol                                       1 mol  

So its clear from the reaction that 2 mole of MnO₂ gives 1 mole of Zn(OH)₂

then how many moles of Zn(OH)₂ will be produce by 0.25 moles of MnO₂

So.

apply unity formula

         2 mol of MnO₂ ≅ 1 mole of Zn(OH)₂

            0.25 moles of MnO₂  ≅ X mole of Zn(OH)₂

Do cross multiplication

           X mole of Zn(OH)₂ = 1 mole x 0.25 mol x  / 2 mol

         X mole of Zn(OH)₂  ≅ 0.125

Now Conver moles of  Zn(OH)₂  to mass

Formula used

         mass in grams = no. of moles x molar mass

Molar mass of  Zn(OH)₂

Molar mass of  Zn(OH)₂ = 65.4 + 2 (16 + 1)

Molar mass of  Zn(OH)₂ = 65.4 + 2 (17)

Molar mass of  Zn(OH)₂ = 65.4 + 34 = 99.4 g/mol

Put values in above equation

        mass in grams = 0.125 mol x 99.4 g/mol  

        mass in grams = 12.43 g

So,

12.43 g of Zn(OH)₂  will be produce.

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