All categories

2 years ago

How will you known that you have reached the equivalence point when titrating the NaOH solution with HCl?

Chemistry

1 answer:

Nitella [24]2 years ago

6 0

Answer:

When the solution (with phenolphthalein) changes to colorless

Explanation:

When titrating with HCl is common to add phenolphthalein as an acid-base indicator.

Phenolphthalein is pink or fucsia when added into a basic solution. On the other hand when it is in acid solutions, is colorless.

So, when titrating, the NaOH solution will be initialy pink due to the phenolphthalein and when reaching the equivalence point, that color will fade out into colorless. This is how you know you hace reached the equivalent point.

You might be interested in

What did Ernest Rutherford s gold foil experiment demonstrate about atoms?

Ghella [55]

-Positively charged nucleus
-Empty spaced
-Dense core

5 0

3 years ago

Ammonium Phosphate (NH4)3PO4 is a strong electrolyte. What will be the concentration of all the ions in a 0.9 M solution of ammo

katrin2010 [14]

There are 3 moles of

$NH_4^+$

per 1 mole of salt and 1 mole of

$PO_{4}^{3-}$

per mole of salt, the total ionic concentrations must be

$2.7 M$ of $NH_4^{+}$ , and $0.9 M$ of $PO_4^{3-}$

3 0

3 years ago

Use this equation for the following problems: 2NaN3 --> 2Na+3N2

olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

2NaN₃ → 2Na + 3N₂,

It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

using cross multiplication:

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

Finally, we can get the grams of NaN₃ needed:

mass = no. of moles x molar mass = (0.517 mol)(65.0 g/mol) = 33.6 g.

Q2: How many mL of N₂ result if 8.3 g Na are also produced?

We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

We can get the no. of moles of N₂ produced with 0.36 mol of Na:

using cross multiplication:

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

We can get the mass of 0.54 mol of N₂:

mass = no. of moles x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

Now, we can get the mL of 15.12 g of N₂:

using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

∴ The volume of N₂ result = (1.0 L)(15.12 g)/(0.92 g) = 16.434 L = 16434 mL.

4 0

2 years ago

Use the Lab Report document as a template. Use the descriptions and examples of each part to help you write your Lab Report. You

GarryVolchara [31]

This link might help you!!

https://www.eiu.edu/biology/bio1500/writing_a_lab_report.pdf

6 0

2 years ago

The net charge on a sulfide ion (s2-) is -2. explain how this ion obtains its charge

Anestetic [448]

Sulfur is a group six element in period 3 with atomic number 16 and an electronic configuration of 2:8:6. Therefore, to attain a stable configuration it requires to gain two electrons forming an ion with a charge of -2. The negative charge is due to the gaining of electrons.

6 0

3 years ago