Titration lab sheet: day 2 (alternate)‼️‼️‼️

See explanation below

Explanation:

As the problem states, this is an acid base titration, and both titrations are already saying that they were both neutralized. When an acid base titration is neutralized, means that it reach it's equivalence point. In this point, we can say that the moles of the acid are the same moles of the base. In other words the following:

n₁ = n₂   (1)

1 is the acid and 2 is the base.

You should note that the above expression is real when the mole ratio is 1:1. When it's not, we need to see the mole ratio and then, adjust the expression to that.

the moles can also be expressed as:

n = M * V

Replacing in the first expression we have:

M₁V₁ = M₂V₂  (2)

With this expression we can calculate either the volume or concentration of the compounds given. Let's do this by parts:

Titration 1:

In this case we have KOH and H₂SO₄, so the balanced reaction would be:

2KOH + H₂SO₄ > K₂SO₄ + 2H₂O

As you can see, we have 2 moles of KOH and 1 mole of the acid, so the mole ratio is 2:1, therefore, expression (2) becomes:

M₁V₁ = 2M₂V₂

From here, we solve for concentration of the acid (M₁)

M₁ = 2M₂V₂ / V₁

Replacing the given values we have:

M₁ = 2 * 25 * 0.15 / 15

Now, how can we fill the chart? Is easy, we just put the obtained values:

For the acid it would be:

Solution: H₂SO₄;   Molar ratio: 1;    Volume: 15 mL;  Concentration: 0.5 M

For the base:

Solution: KOH;   Molar ratio: 2;    Volume: 25 mL;    Concentration: 0.15 M

Titration 2:

In this case we do the same thing as before but with different data. First the equation:

HBr + NaOH > NaBr + H₂O

The equation is already balanced and we can see a mole ratio of 1:1, so we can use expression (2) and solve for concentration of the base instead:

M₁V₁ = M₂V₂

M₂ = M₁V₁ / V₂

M₂ = 30 * 0.250 / 20

The chart can be filled the same way as in titration 1:

For the acid it would be:

Solution: HBr;   Molar ratio: 1;    Volume: 30 mL;  Concentration: 0.25 M

For the base:

Solution: NaOH;   Molar ratio: 1;    Volume: 20 mL;    Concentration: 0.375 M

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).