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harina [27]
3 years ago
12

Calculate the pOH and pH of a solution which contains 0.001 M NaOH. Assume 100% ionization. (Need an in-depth explanation with f

ormulas please)
Chemistry
1 answer:
Studentka2010 [4]3 years ago
8 0

Answer:

pH: 11

pOH: 3

Explanation:

NaOH is a strong base which means that it dissociates completely in water. It will break apart into Na⁺ and OH⁻ ions.

NaOH ⇒ Na⁺ + OH⁻

Because NaOH dissociates completely into its respective ions in water, the moles of NaOH is equal to the moles of hydroxide ions. So, [OH⁻] = 0.001 M.

Now to find the pOH, use the formula pOH = -log[OH⁻].

pOH = -log[OH⁻]

= -log(0.001)

= 3

The pOH of the solution is 3.

To find the pH, subtract the pOH from 14 since pH + pOH = 14.

14 - 3 = 11

The pH of the solution is 11.

Hope that helps.

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How many moles are in 297 g of NH3?
BigorU [14]

Answer:

1. 17.4 moles.

2. 1.13 moles

3. 315.5 moles

4. 390.6g

5. 1.13x10⁶ moles

6. 14.8 moles

7. 337 moles

8. 2.15x10²⁴ molecules

9. 3.13x10²⁴ atoms

10. 1.38x10²⁴ particles

11. 1517g

12. 455g of CaF₂

Explanation:

We can convert formula units to moles or vice versa using Avogadro's number and moles to grams using molar mass of the substance:

1. Molar mass NH3: 17.031g/mol

297g * (1mol / 17.031g) = 17.4 moles

2. Molar mass MgCO3: 84.3g/mol

95g * (1mol / 84.3g) = 1.13 moles

3. Using Avogadro's number (6.022x10²³ formula units / mol):

1.9x10²⁶FU * (1mol / 6.022x10²³FU) = 315.5 moles

4. Molar mass H2O: 18g/mol

21.7mol * (18g / mol) = 390.6g

5. Using Avogadro's number (6.022x10²³ molecules / mol):

6.78x10²⁹molecules * (1mol / 6.022x10²³FU) = 1.13x10⁶ moles

6. 8.9x10²⁴FU * (1mol / 6.022x10²³FU) = 14.8 moles

7. Using Avogadro's number (6.022x10²³ atoms / mol):

2.03x10²⁶atoms* (1mol / 6.022x10²³FU) = 337 moles

8. 3.569mol * (6.022x10²³ molecules / 1mol) = 2.15x10²⁴ molecules

9. 5.2mol * (6.022x10²³ atoms / 1mol) = 3.13x10²⁴ atoms

10. Molar mass Li₂SO₄: 109.94g/mol:

36g * (1mol / 109.94g) * (6.022x10²³ molecules / 1mol) * (7 particles / 1molecule) = 1.38x10²⁴ particles

<em>Assuming particles are atoms and in 1 molecule of Li₂SO₄ you have 7 atoms.</em>

11. Molar mass Cl₂: 70.9g/mol:

21.4mol * (70.9g / mol) = 1517g

12. Molar mass CaF₂: 78.07g/mol:

3.51x10²⁴FU * (1mol / 6.022x10²³FU) * (78.07g / mol) = 455g of CaF₂

8 0
3 years ago
HELP PLS (CHEMISTRY)
IRINA_888 [86]

The first most obvious thing to note is when naming transitional metals, you have to state its charge with roman numerals (except for 1 if I remember correctly). For example, Iron (lll), iron has a charge of 3.

4 0
2 years ago
Ten lab safety rules
Bogdan [553]
Don't touch your eyes.

Never taste-test unless the teacher tells you to.

Do not touch anything without directions.

Wear safety goggles. 

Wash your hands after each experiment.

Wear proper lab clothes.

Do not mishandle lab equipment.

Clean up your workplace.

Act serious; no horseplay!

Report accidents to the teacher right away!


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4 0
3 years ago
The options are: ( it can’t be repeated )
Artyom0805 [142]

Answer:

3- gamma radiation

Explanation:

Hello,

In the above question, 4 of the options are related to polymerization which are

1. Synthetic polymer

2. Natural polymer

3. Condensation polymerization

4. Addition polymerization.

The first two options are types of polymer that exists while the last two are polymerization techniques.

The odd option here which is "gamma radiation" is a particle which is emitted from radioactive substances during decay. It has no mass and no charge but it is highly penetrating and dangerous to human health.

However,

Synthetic polymers are also known as man made polymers and they exist around us because they're present in materials which we use everyday. An example is polyethylene, nylon-6,6 etc

Natural polymers are compounds which are polymeric in nature (compounds catenating to form a complex molecule). Natrual occurring polymers can be found in proteins and some lipids.

4 0
3 years ago
What is the correct formula name for calcium chloride?
Leokris [45]
C. CaCl2 is the correct answer
7 0
3 years ago
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