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mihalych1998 [28]
4 years ago
7

Why is it important that acids and bases have certain naming rules? Why shouldn't we just call them whatever we want?​

Chemistry
2 answers:
DanielleElmas [232]4 years ago
7 0

Explanation:

This naming can help identify a solution as acid or base (and some of its properties) even before conducting tests. It also allows scientists across the world to collaborate through using a standard naming system.

Naming of the acids and bases is significant in identifying the type and properties of the acid or base. Acids, for example, are named based on the anion attached to the H+ ion in the chemical structure. For example, HCl is called hydrochloric acid because the H+ is bound to a Chloride ion. The suffix for acid is;  –ic, -ate & -ous.

Bases, on the other hand, are named based on the cation attached to the hydroxide eg, CaOH is named calcium hydroxide. The suffix for bases is -ide.

maks197457 [2]4 years ago
6 0

Answer:

<em>It is important to have certain naming rules for both acids or bases so we can distinguish them from one another. </em>

Explanation:

These scientific names are used all over the world. Without these rules, one would not be able to tell the difference between an acid or base. These names not only explain the nature of these substances as well as their reactive properties.  

<u>The name of an acid is classified on the basis of its anions while a base is named using the ionic system. </u>

The examples  of common bases are <em>potassium hydroxide, sodium hydroxide </em>etc while common acids are <em>sulfuric acid, citric acid</em> etc.

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Answer:

Molality = 1.46 molal

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Explanation:

Step 1: Data given

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Mass urea = 13.40 grams

Molar mass urea = 60.1 g/mol

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Step 2: Calculate moles urea

Moles urea = mass urea /molar mass urea

Moles urea = 13.40 grams / 60.1 g/mol

Moles urea = 0.223 moles

Step 3: Calculate the molality

Molality = moles urea / mass water

Molality = 0.223 moles / 0.1532 kg

Molality = 1.46 molal

Step 4: Calculate the freezing point of the solution

ΔT = i * Kf * m

ΔT = 1* 1.86 °C/m * 1.46 m

ΔT = 2.72 °C

The freezing point = -2.72 °C

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