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melomori [17]
2 years ago
8

250 kJ of energy are removed from a 4.00 x 102 g sample of water at 60˚C. Will the sample of water completely freeze?

Chemistry
1 answer:
Stolb23 [73]2 years ago
5 0

250 kJ of energy are removed from a 4.00 x 102 g sample of water at 60˚C. Will the sample of water completely freeze: Yes, because there is enough energy.

<h3>At what temperature would a sample of water freeze?</h3>
  • Note from the Facilitator: At certain temperatures, water changes its condition due to temperature variations. At sea level, fresh water changes from a solid to a liquid at 32°F (0°C). Liquid water freezes at temperatures below 32°F (0°C); this temperature is known as the freezing point of water.
  • The fact that a single water molecule cannot transform into a solid, liquid, or gas is the answer. These names refer to collective behaviors of water molecules rather than to individual molecules.
  • For instance, the solid (ice) has a collection of molecules that are bound together and arranged in a predictable manner. That cannot be accomplished by a single molecule alone

250 kJ of energy are removed from a 4.00 x 102 g sample of water at 60˚C. Will the sample of water completely freeze: Yes, because there is enough energy.

To learn more about water freezing, refer to:

brainly.com/question/15209660

#SPJ9

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An alkaline battery produces electrical energy according to the following equation.
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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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