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hjlf
3 years ago
9

_________ is a systematic method of naming chemical compounds. This creates an unambiguous and consistent name for any chemical

compound throughout the world. Example: sodium chloride - NaCl hydrochloric acid - HCl carbon tetrachloride - CCl4
Chemistry
1 answer:
Vlada [557]3 years ago
6 0

Answer: IUPAC NOMENCLATURE

Explanation:

IUPAC stands for International Union of Pure and Applied Chemistry. They devised a systematic method for naming compounds in order to create a uniform global unambiguous system of nomenclature hence making it easier for researchers to share information more freely without the hindrance of reporting the same compound using different names in different parts of the world thus creating confusion in chemical literature.

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Detrivores; i.e, Earth Worms, Rolly Pollies etc.
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What is the molarity of a solution containing 0.32 moles of NaCI in 3.4 liters
CaHeK987 [17]

Answer:

.094 M

Explanation: 0.32molNaCl/ 3.4L = .094M

3 0
3 years ago
A 27.9 mL sample of 0.289 M dimethylamine, (CH3)2NH, is titrated with 0.286 M hydrobromic acid.
sesenic [268]

Answer:

(1) Before the addition of any HBr, the pH is 12.02

(2) After adding 12.0 mL of HBr, the pH is 10.86

(3) At the titration midpoint, the pH is 10.73

(4) At the equivalence point, the pH is 5.79

(5) After adding 45.1 mL of HBr, the pH is 1.18

Explanation:

First of all, we have a weak base:

  • 0 mL of HBr is added

(CH₃)₂NH  + H₂O  ⇄  (CH₃)₂NH₂⁺  +  OH⁻            Kb = 5.4×10⁻⁴

0.289 - x                             x                x

Kb = x² / 0.289-x

Kb . 0.289 - Kbx - x²

1.56×10⁻⁴ - 5.4×10⁻⁴x - x²

After the quadratic equation is solved x = 0.01222 → [OH⁻]

- log  [OH⁻] = pOH → 1.91

pH = 12.02   (14 - pOH)

  • After adding 12 mL of HBr

We determine the mmoles of H⁺, we add:

0.286 M . 12 mL = 3.432 mmol

We determine the mmoles of base⁻, we have

27.9 mL . 0.289 M = 8.0631 mmol

When the base, react to the protons, we have the protonated base plus water (neutralization reaction)

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       3.432 mm                 -

4.6311 mm                                  3.432 mm

We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.

[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M

[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M

We have just made a buffer.

pH = pKa + log (CH₃)₂NH  / (CH₃)₂NH₂⁺

pH = 10.73 + log (0.116/0.0860) = 10.86

  • Equivalence point

mmoles of base = mmoles of acid

Let's find out the volume

0.289 M . 27.9 mL = 0.286 M . volume

volume in Eq. point = 28.2 mL

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       8.0631mm               -

                                                8.0631 mm

We do not have base and protons, we only have the conjugate acid

We calculate the new concentration:

mmoles of conjugated acid / Total volume (initial + eq. point)

[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL)  = 0.144 M

(CH₃)₂NH₂⁺   +  H₂O   ⇄   (CH₃)₂NH  +  H₃O⁻       Ka = 1.85×10⁻¹¹

 0.144 - x                                  x               x

[H₃O⁺] = √ (Ka . 0.144) →  1.63×10⁻⁶ M  

pH = - log [H₃O⁺] = 5.79

  • Titration midpoint (28.2 mL/2)

This is the point where we add, the half of acid. (14.1 mL)

This is still a buffer area.

mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)

mmoles of base = 8.0631 mmol - 4.0326 mmol

[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M

[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M

pH = pKa + log (0.096M / 0.096 M)

pH = 10.73 + log 1 =  10.73

Both concentrations are the same, so pH = pKa. This is the  maximum buffering capacity.

  • When we add 45.1 mL of HBr

mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol

mmoles of base = 8.0631 mmoles

This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm     12.8986 mm             -

       -               4.8355 mm                        

[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)

[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M

- log [H₃O⁺] = pH

- log 0.0662 = 1.18 → pH

7 0
3 years ago
Describe three characteristics of the outer core and explain why the outer core is so important to humans on earth.
iren2701 [21]

Answer:

The outer core is the only layer that is liquid. It is also mainly made from nickel and iron. The main job of the outer core is that it's responsible for the Earth's magnetic field. As the earth spins, the liquid inside this layer spins aswell, keeping it balanced.

6 0
2 years ago
Given the following chemical reaction: 2 O2 + CH4 CO2 + 2 H2O What mass of CH4 is required to completely react with 100 grams of
AfilCa [17]

Answer:

The mass methane (CH4) required is, 25 grams  (option C)

Explanation:

Mass of oxygen gas = 100 g

Molar mass of oxygen gas = 32 g/mole

Molar mass of methane gas = 16 g/mole

<u>Step 1:</u> Balance the reaction

2O2 + CH4 → CO2 + 2H2O

<u>Step 2:</u> Calculate the moles of O2:

Moles O2 = mass of O2 / Molar mass of o2  

Moles O2 = 100g / (32g/mole) = 3,125 moles

⇒From the balanced reaction we conclude that  2 moles of O2 react with 1 mole of CH4

⇒ So, 3.125 moles of O2 react with 3.125/2 = 1.563 moles of CH4

<u>Step 3:</u> Calculate the mass of CH4:

Mass of CH4 = moles of CH4 x Molar mass of CH4

Mass of CH4 = 1.563 moles / (16g/ moles) = 25.008 grams CH4

6 0
3 years ago
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