How can you tell just by looking whether a mixture in water is a solution or a suspension?

Answer:

Yes

Explanation:

Yes, you can test this by comparing grades you received when you studied and when you did not study

Answer:

86.04 %

Explanation:

Data given:

mass of Al  = 160.0 g

actual yield Al₂O₃= 260 g

Theoretical yield = ?

percent yield ofAl₂O₃ = ?

Solution:

First we have to find theoretical yield.

So,

we Look at the balanced reaction

             4Al  + 3O₂ -----—> 2Al₂O₃

             4 mol                      2 mol

As 4 mole of Al give 2 mole of Al₂O₃

Convert moles to mass

molar mass of Al = 27 g/mol

molar mass of Al₂O₃ = 2(27) + 3(16)

molar mass of Al₂O₃ = 54 + 48

molar mass of Al₂O₃ = 102 g/mol

Now

             4Al         +       3O₂   -----—>     2Al₂O₃

        4 mol (27g/mol)                       2 mol (102 g/mol)

             108 g                                            204 g

108 grams of Al produce 204 g of Al₂O₃

So

if 108 grams of Al produce 204 g of Al₂O₃ so how many grams of Al₂O₃will be produced by 160 g of Al.

Apply Unity Formula

              108 grams of Al  ≅ 204 g of Al₂O₃

              160 grams of Al ≅ X of Al₂O₃

Do cross multiply

              mass of Al₂O₃= 204 g x 160 g / 108 g

              mass of Al₂O₃ = 302.2 g

So the Theoretical yield of Al₂O₃ = 302.2 g

Now Find the percent yield of Al₂O₃

Formula Used

             percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

            percent yield = 260g / 302.2 g x 100 %

            percent yield = 86.04 %

percent yield of Al₂O₃ = 86.04 %