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inysia [295]
2 years ago
9

What are the prefixes for molecular compounds?

Chemistry
2 answers:
fredd [130]2 years ago
8 0

Answer:

In nomenclature of simple molecular compounds, the more electropositive atom is written first and the more electronegative element is written last with an -ide suffix.

The Greek prefixes are used to dictate the number of a given element present in a molecular compound.

Prefixes can be shortened when the ending vowel of the prefix “conflicts” with a starting vowel in the compound.

Common exceptions exist for naming molecular compounds, where trivial or common names are used instead of systematic names, such as ammonia (NH3) instead of nitrogen trihydride or water (H2O) instead of dihydrogen monooxide.

Terms

nomenclatureA set of rules used for forming the names or terms in a particular field of arts or sciences.

electronegativeTending to attract electrons within a chemical bond.

electropositiveTending to not attract electrons (repel) within a chemical bond.

Chemical Nomenclature

The primary function of chemical nomenclature is to ensure that a spoken or written chemical name leaves no ambiguity concerning to what chemical compound the name refers. Each chemical name should refer to a single substance. Today, scientists often refer to chemicals by their common names: for example, water is not often called dihydrogen oxide. However, it is important to be able to recognize and name all chemicals in a standardized way. The most widely accepted format for nomenclature has been established by IUPAC.

Molecular compounds are made when two or more elements share electrons in a covalent bond to connect the elements. Typically, non-metals tend to share electrons, make covalent bonds, and thus, form molecular compounds.

Rules for Naming Molecular Compounds:

Remove the ending of the second element, and add “ide” just like in ionic compounds.

When naming molecular compounds prefixes are used to dictate the number of a given element present in the compound. ” mono-” indicates one, “di-” indicates two, “tri-” is three, “tetra-” is four, “penta-” is five, and “hexa-” is six, “hepta-” is seven, “octo-” is eight, “nona-” is nine, and “deca” is ten.

If there is only one of the first element, you can drop the prefix. For example, CO is carbon monoxide, not monocarbon monoxide.

If there are two vowels in a row that sound the same once the prefix is added (they “conflict”), the extra vowel on the end of the prefix is removed. For example, one oxygen would be monooxide, but instead it’s monoxide. The extra o is dropped.

Generally, the more electropositive atom is written first, followed by the more electronegative atom with an appropriate suffix. For example, H2O (water) can be called dihydrogen monoxide (though it’s not usually). Organic molecules (molecules made of C and H along with other elements) do not follow this rule.

sergiy2304 [10]2 years ago
4 0
<h2>Answer:</h2>

Here are the prefixes in naming molecular compounds:

Mono- 1

Di- 2

Tri- 3

Tetra- 4

Penta- 5

Hexa- 6

Hepta- 7

Octa- 8

Nona- 9

Deca- 10

Molecular compounds are named using a systematic approach of prefixes to indicate the number of each element present in the compound.

<em>I</em><em> </em><em>h</em><em>o</em><em>p</em><em>e</em><em> </em><em>i</em><em>t</em><em> </em><em>h</em><em>e</em><em>l</em><em>p</em><em>s</em><em> </em>●~●

<h3>#CarryOnLearning</h3>
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Answer:

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8 0
3 years ago
50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?
yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

           = 25.00 x 0.200

           = 5.00 m-mol

           = 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

                                           = 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

                                                        = 0.167 M

Now the ICE table :

            $NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M)       0.167                   0            0

C (M)         -x                      +x          +x

E (M)      0.167-x                  x           x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$            $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

    = $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

       = $- \log(1.9054 \times 10^{-6} \ M)$

        =5.72

Now, since pH + pOH = 14

           pH = 14.00 - 5.72

                = 8.279

Therefore the ph is 8.279 at the end of the titration.

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