how many protons electrons and neutrons does a atom with the atomic number 50 and mass number 125 contain?

Answer:

1. Temperature

2. Surface area

3. Catalyst

Explanation:

Answer:

Heat, molecular size, and surface area

Explanation:

Heat: higher temperatures are directly correlated to higher kinetic energy. With more kinetic energy, the solute molecules move faster and the bonds are more likely to break.

Molecule size: usually at the same temperature and pressure, the solute with smaller molecular size dissolves faster. A large molecule usually has a heavier weight and size, which makes it more difficult for the solvent to surround it and help break its bonds.

Surface area: increasing surface area generally increases the solubility rate. This is because with a higher area, more molecules are exposed to the solvent, so their bonds are more likely to break.

Answer: The correct answer is Option b.

Explanation:

Double displacement reactions are defined as the reactions in which exchange of ions takes place.

For the given options:

Option a:

This is an example of synthesis reaction.

Option b:

This is an example of double displacement reaction.

Option c:

This is an example of decomposition reaction.

Option d:

This is an example of single displacement reaction.

From the above information, the correct answer is Option b.

Answer:

Part 1

Theoretical yield of MgO for trial 1 = 0.84 g

Theoretical yield of MgO for trial 2 = 1.01 g

Part 2

Percent yield trial 1 = 28.6 %

Percent yield trial 2 = 49.9 %

Part 3

Average percent yield of MgO for two trial = 39.25 %

Explanation:

Part 1.

Data Given

                                                              Trial 1                     Trial 2

mass of empty crucible and lid:          26.679 g               26.685 g

mass of Mg metal, crucible and lid:    26.931 g               26.988 g

mass of MgO, crucible and lid:            27.090 g              27.179 g

Theoretical yield of MgO for trial 1 and 2 = ?

Solution:

As Mg is limiting reagent so amount of MgO depends on the amount of Mg.

So, now we will look for the reaction to calculate theoretical yield

MgO form by the following reaction:

               Mg  +  O₂ --------->  2 MgO

              1 mol                        2 mol

Convert moles to mass

Molar mass of Mg = 24 g/mol

Molar mass of MgO = 24 + 16 = 40 g/mol

So,

                     Mg        +         O₂      --------->     2 MgO

            1 mol (24 g/mol)                                  2 mol(40 g/mol)

                   24 g                                                    80 g

So,

24 g of Mg gives 80 g of MgO

To Calculate theoretical yield of MgO for Trial 1

First we look for the mass of Mg in the Crucible

• mass of Mg for trial 1

Mass of Mg = mass of Mg metal, crucible and lid - mass of empty crucible and lid

       Mass of Mg = 26.931 g - 26.679 g

       Mass of Mg = 0.252 g

As we come to know that 24 g of Mg gives 80 g of MgO, then amount of Mg from trial 1 that is 0.252 g will produce how many grams of MgO

Apply unity formula

               24 g of Mg ≅ 80 g of MgO

               0.252 g of Mg ≅ X g of MgO

Do cross multiplication

               X g of MgO = 0.252 g x 80 g / 24 g

               X g of MgO = 0.84 g

So the theoretical yield of MgO is  0.84 g

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To Calculate theoretical yield of MgO for Trial 2

First we look for the mass of Mg in the Crucible

• mass of Mg for trial 2

Mass of Mg = mass of Mg metal, crucible and lid - mass of empty crucible and lid

      Mass of Mg = 26.988 g - 26.685 g

      Mass of Mg = 0.303 g

As we come to know that 24 g of Mg gives 80 g of MgO, then amount of Mg from trial 2 that is 0.303 g will produce how many grams of MgO

Apply unity formula

               24 g of Mg ≅ 80 g of MgO

                0.303 g of Mg ≅ X g of MgO

Do cross multiplication

               X g of MgO = 0.303 g x 80 g / 24 g

               X g of MgO = 1.01 g

So the theoretical yield of MgO is  1.01 g

__________________________

Part 2

percent yield of MgO for trial 1 and 2 = ?

Solution:

For trial 1

To calculate percent yield we have to know about actual yield of MgO

• mass of MgO for trial 1

Mass of MgO = mass of MgO, crucible and lid - mass of empty crucible and lid

    Mass of MgO =  27.090 g -  26.685 g

    Mass of MgO =  0.24 g

And we also know that

Theoretical yield of MgO for trial 1 = 0.84 g

Formula used

       Percent yield = actual yield / theoretical yield x 100

put values in above formula

       Percent yield =  0.24 g / 0.84 g x 100

       Percent yield = 28.6 %

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For trial 2

To calculate percent yield we have to know about actual yield of MgO

• mass of MgO for trial 2

Mass of MgO = mass of MgO, crucible and lid - mass of empty crucible and lid

    Mass of MgO =  27.179 g -  26.685 g

    Mass of MgO =  0.494 g

And we also know that

Theoretical yield of MgO for trial 2 = 1.01 g

Formula used

       Percent yield = actual yield / theoretical yield x 100

put values in above formula

       Percent yield =   0.494 g/ 1.01 g x 100

       Percent yield = 49.9 %

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Part 3

average percent yield of MgO for the two trials =?

Solution:

As we know

Percent yield trial 2 = 28.6 %

Percent yield trial 2 = 49.9 %

Formula used

Average percent yield = percent yield trial 1 + percent yield trial 2 / 2

Put values in above formula

           Average percent yield = 28.6 + 49.9  / 2

           Average percent yield = 78.5 / 2

           Average percent yield = 39.25 %

Average percent yield of MgO for two trial = 39.25 %